Solved Examples

How to Use This Problem Bank

Eleven sections of theory, and now the part that actually earns marks. What follows is 45 worked problems, arranged easy first and hard last, covering the whole chapter — from "which formula is this?" up to the multi-step cycles that decide ranks.

Work them with a pen and paper. Cover the solution, try it, then compare — including the check at the end of each one, because the check is where the marks usually leak away.

The Sign Convention

Key Point — THIS CHAPTER'S SIGN CONVENTION:

• $\Delta Q$ is positive when heat is added TO the system, negative when heat leaves it.
• $\Delta W$ is positive when work is done BY the system (expansion), negative when work is done ON it (compression).
• $\Delta U$ is the change in internal energy, a state function.

The first law is then $$\Delta Q = \Delta U + \Delta W$$

Your chemistry course, and many physics books, write $\Delta U = Q + W$ with $W$ meaning the work done ON the system. That is the same physics with one sign moved, and this section mentions it here and then never again. Every solution below uses $\Delta Q = \Delta U + \Delta W$ exclusively, and every solution states the sign of each of the three quantities and says what that sign means physically. A bare number is not an answer in this chapter.

The four questions to ask before you write anything

1. Is the temperature in kelvin? Every ratio $\frac{T_2}{T_1}$, every gas-law step, every efficiency and every adiabatic relation needs absolute temperature. Only a bare difference $\Delta T$ is the same number in Celsius and kelvin. This is the commonest wrong answer in the chapter.
2. Which process is it? Constant $T$, constant $Q$, constant $P$ or constant $V$? Name it before you reach for a formula — three of the four kill a whole term of the first law outright.
3. Is the gas expanding or being compressed? That fixes the sign of $\Delta W$ before any arithmetic. Volume up means $\Delta W > 0$.
4. Is this one process, or a closed loop? Round a loop $\Delta U = 0$ exactly, so $\Delta Q = \Delta W$ and the net work is the enclosed area. Spotting a loop turns a hard question into a table.

Key Point — every formula this section uses, in one place: $$\Delta Q = \Delta U + \Delta W, \qquad dQ = dU + P\,dV, \qquad W = \int_{V_1}^{V_2} P\,dV$$ $$C_p - C_v = R, \qquad \gamma = \frac{C_p}{C_v}, \qquad C_v = \frac{R}{\gamma - 1}, \qquad C_p = \frac{\gamma R}{\gamma - 1}$$ $$\Delta U = nC_v\,\Delta T \ \text{(any process, ideal gas)}, \qquad \Delta Q_V = nC_v\Delta T, \qquad \Delta Q_P = nC_p\Delta T$$ $$W_{\text{isothermal}} = nRT\ln\frac{V_2}{V_1} = nRT\ln\frac{P_1}{P_2}, \qquad W_{\text{isobaric}} = P(V_2 - V_1) = nR\,\Delta T$$ $$PV^{\gamma} = \text{const}, \qquad TV^{\gamma-1} = \text{const}, \qquad P^{1-\gamma}T^{\gamma} = \text{const}$$ $$W_{\text{adiabatic}} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$ $$\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1}, \qquad \alpha = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}, \qquad \alpha_{\text{hp}} = \frac{Q_1}{W} = \alpha + 1$$ $$\eta_{\text{Carnot}} = 1 - \frac{T_2}{T_1}, \qquad \frac{Q_1}{Q_2} = \frac{T_1}{T_2}, \qquad \alpha_{\text{Carnot}} = \frac{T_2}{T_1 - T_2}$$

A note on symbols, once

$T$ is always an absolute temperature in kelvin; $t$ or $t_C$ is Celsius, and $T = t_C + 273.15$. $T_1$ is the source (hot) and $T_2$ is the sink (cold) — also written $T_H$ and $T_C$ — and $Q_1$ is the heat drawn from the source while $Q_2$ is the heat rejected to the sink, both quoted as magnitudes with the direction stated in words. $n$ is the number of moles, also written $\mu$. $C_p$ and $C_v$ are molar specific heats in J/(mol K); lowercase $c_p$ and $c_v$ are per kilogram, in J/(kg K). $\eta$ is efficiency and $\alpha$ is the coefficient of performance, which is never called an efficiency because it routinely exceeds 1.

The constants used below

Every solution also restates the constants it uses inside itself, so you never have to scroll back.

Quantity	Value used
Gas constant $R$	8.314 J/(mol K)
1 atmosphere	$1.013 \times 10^{5}$ Pa
Absolute zero offset	$0°$C $= 273.15$ K
1 calorie	4.186 J
$\gamma$, monatomic (He, Ne, Ar)	$\frac{5}{3} \approx 1.667$
$\gamma$, diatomic (H$_2$, N$_2$, O$_2$, air)	$\frac{7}{5} = 1.400$
$\gamma$, polyatomic (CO$_2$, NH$_3$)	$\frac{4}{3} \approx 1.33$
$C_v$ monatomic / diatomic / polyatomic	12.47 / 20.79 / 24.94 J/(mol K)
$C_p$ monatomic / diatomic / polyatomic	20.79 / 29.10 / 33.26 J/(mol K)
Specific heat capacity of water	4186 J/(kg K)

The values of $\gamma$, and the fact that work is the area under a $P$-$V$ curve, sit outside the rationalised syllabus body text, yet nothing here can be solved without them — Boards, JEE Main and NEET all ask about them every year, so they are used freely throughout.

[Board Important] Every solution writes the formula on its own line before any number goes into it, and every temperature that enters a ratio is written with the letter K next to it. Do both in the exam. A correct formula with an arithmetic slip still earns most of the marks; a Celsius value inside $\frac{T_2}{T_1}$ loses them all.

Solved Examples

Part 1: Systems, State Variables and the Zeroth Law

Four warm-ups. No first law yet — just the vocabulary, used precisely. Everything in this part is worth exactly the marks that a careless reader throws away.

Key Point: An open system exchanges both matter and energy with its surroundings; a closed system exchanges energy only; an isolated system exchanges neither. The classification depends on the boundary you choose, not on the object — which is why the same kitchen pot can be all three in the space of ten minutes.

Example 1: A pressure cooker at three moments

A pressure cooker sits on a flame. Classify the gas-plus-water inside as open, closed or isolated at each of these moments: (a) the lid is off and the water is boiling; (b) the lid is sealed, the weight is down, and the flame is on; (c) the sealed cooker is taken off the flame and wrapped in a thick blanket of expanded polystyrene.

Solution:

1. Fix the system and the boundary first. Take the system to be everything inside the cooker: the water and the steam above it. The boundary is the inner surface of the pot plus, when there is one, the lid.

2. (a) Lid off, boiling. Steam is leaving through the open top, so matter crosses the boundary. Heat is arriving from the flame, so energy crosses too. Both cross: this is an open system. Notice that the mass of the system is falling, which is exactly why an open system cannot be described by a fixed $n$ in $PV = nRT$.

3. (b) Lid sealed, weight down, flame on. Nothing can get past the sealed lid while the weight stays seated, so no matter crosses. But the metal base conducts heat in from the flame, so energy still crosses. Matter no, energy yes: a closed system. This is the case almost every thermodynamics problem means when it says "a gas in a cylinder".

4. (c) Off the flame, wrapped in insulation. No matter crosses, and now the insulation blocks heat as well. Nothing crosses: an isolated system — to the extent that the insulation is perfect, which it never quite is.

5. The catch worth noticing. The cooker in (c) is not doing any work either, so its internal energy $U$ is now fixed. That is what "isolated" buys you: a constant $U$, which is the strongest constraint in the whole subject.

Final Answer: (a) open, (b) closed, (c) isolated.

Takeaway: A system is open, closed or isolated because of its boundary, not because of what it is made of. Draw the boundary explicitly before you classify anything, and re-draw it whenever the physical setup changes.

Example 2: Turning extensive quantities into intensive ones

A rigid cylinder holds 2.00 moles of helium (molar mass 4.00 g/mol) in a volume of $0.0500$ m$^3$ at 300 K. Find its mass, pressure, density, molar volume, internal energy and internal energy per kilogram. Then cut the whole thing in half with an imaginary partition and say which of these six numbers changes.

Solution:

1. The constants. $R = 8.314$ J/(mol K); for a monatomic gas $C_v = \frac{3}{2}R = 12.471$ J/(mol K). The temperature is already absolute: 300 K.

2. Mass and pressure. $$m = nM = 2.00 \times 4.00 = 8.00 \text{ g} = 8.00 \times 10^{-3} \text{ kg}$$ $$P = \frac{nRT}{V} = \frac{2.00 \times 8.314 \times 300}{0.0500} = 9.98 \times 10^{4} \text{ Pa}$$

3. Internal energy. For an ideal gas $U = nC_vT$, and for helium $C_v = 12.471$ J/(mol K): $$U = 2.00 \times 12.471 \times 300 = 7482.6 \text{ J}$$

4. The three ratios. $$\rho = \frac{m}{V} = \frac{8.00\times10^{-3}}{0.0500} = 0.160 \text{ kg/m}^3$$ $$V_m = \frac{V}{n} = \frac{0.0500}{2.00} = 0.0250 \text{ m}^3\text{/mol}$$ $$u = \frac{U}{m} = \frac{7482.6}{8.00\times10^{-3}} = 9.353 \times 10^{5} \text{ J/kg}$$

5. Now halve it. Slice the cylinder down the middle. Each half contains 1.00 mole in $0.0250$ m$^3$, of mass $4.00$ g, with internal energy $3741.3$ J. So $n$, $m$, $V$ and $U$ all halve — these are the extensive quantities. But $$\rho = \frac{4.00\times10^{-3}}{0.0250} = 0.160 \text{ kg/m}^3, \qquad V_m = 0.0250 \text{ m}^3\text{/mol}, \qquad u = 9.353\times10^{5} \text{ J/kg}$$ are all unchanged, and so are $P$ and $T$. These are the intensive quantities.

6. The pattern behind it. Every one of the unchanged quantities is a ratio of two extensive quantities. Divide an extensive quantity by another extensive quantity and the size of the system cancels out. That is the entire recipe for manufacturing an intensive variable, and it is why density, molar volume and specific internal energy exist at all.

Final Answer: $m = 8.00$ g, $P = 9.98\times10^{4}$ Pa, $\rho = 0.160$ kg/m$^3$, $V_m = 0.0250$ m$^3$/mol, $U = 7482.6$ J, $u = 9.353\times10^{5}$ J/kg. Halving the system halves $n$, $m$, $V$ and $U$ and leaves $P$, $T$, $\rho$, $V_m$ and $u$ alone.

Takeaway: Halve the system in your head — whatever halves is extensive, whatever survives is intensive. It is a five-second test and it never fails.

Example 3: Three flasks and one thermometer

Three sealed flasks $A$, $B$ and $C$ stand on a bench. A thermometer is dipped into $C$ and settles at 310.2 K. $A$ is then placed in thermal contact with $C$ and, after a long wait, no heat flows between them. $B$ is separately placed in contact with $C$, and again no net heat flows. What can you say about $A$ and $B$? And what would happen if $A$ and $B$ were brought into contact with each other, with the thermometer never having touched either?

Solution:

1. Write down what "no net heat flows" means. Two bodies in contact with no net heat flow between them are in thermal equilibrium. So $A$ and $C$ are in thermal equilibrium, and $B$ and $C$ are in thermal equilibrium.

2. Apply the zeroth law. The zeroth law of thermodynamics states: if $A$ is in thermal equilibrium with $C$, and $B$ is in thermal equilibrium with $C$, then $A$ and $B$ are in thermal equilibrium with each other. So bringing $A$ and $B$ together will produce no net heat flow, even though they have never met.

3. Now the temperature. The thermometer read 310.2 K in $C$. The thermometer, being in equilibrium with $C$, is at the temperature of $C$. Since $A$ and $B$ are each in equilibrium with $C$, all four objects share one number: $$T_A = T_B = T_C = 310.2 \text{ K}$$ which is $310.2 - 273.15 = 37.05°$C, a healthy body temperature.

4. Why this deserves to be called a law. It sounds like arithmetic — "if $A = C$ and $B = C$ then $A = B$" — but it is a claim about nature, not about algebra. It says that thermal equilibrium is a transitive relation, and nothing forced that to be true. Because it is true, a single number can be attached to a body and called its temperature, and a thermometer calibrated once will work on anything. Without transitivity, a thermometer would be useless: it would tell you about the thermometer's relationship with one particular body and nothing more.

5. The practical reading. The whole of clinical thermometry rests on this. Nobody compares your forehead directly with a reference block of metal in a laboratory in Paris. The thermometer is the third body $C$, and the zeroth law is the only reason its reading means anything.

Final Answer: $A$ and $B$ are in thermal equilibrium with each other and both are at 310.2 K, or $37.05°$C. Placing them in contact produces no net heat flow.

Takeaway: The zeroth law is what makes a thermometer possible. It is not a triviality about equals signs; it is the empirical fact that thermal equilibrium is transitive, and the entire concept of temperature is built on it.

Example 4: Fixing a state with two numbers and reading off the third

A closed cylinder holds $0.400$ mol of nitrogen (a diatomic gas) at a pressure of $2.00 \times 10^{5}$ Pa and a temperature of 350 K. Find the volume and the internal energy. Then decide which of these five quantities are state variables: pressure, volume, temperature, internal energy, and the heat that was supplied to get the gas here.

Solution:

1. The constants. $R = 8.314$ J/(mol K); for a diatomic gas $C_v = \frac{5}{2}R = 20.785$ J/(mol K). The temperature 350 K is already absolute, so it can go straight into the gas law.

2. Volume from the equation of state. $$PV = nRT \qquad \Longrightarrow \qquad V = \frac{nRT}{P} = \frac{0.400 \times 8.314 \times 350}{2.00\times10^{5}}$$ $$V = \frac{1163.96}{2.00\times10^{5}} = 5.82 \times 10^{-3} \text{ m}^3 = 5.82 \text{ L}$$

3. Internal energy. For an ideal gas $U$ depends on temperature alone: $$U = nC_vT = 0.400 \times 20.785 \times 350 = 2909.9 \text{ J}$$

4. Notice what just happened. You were handed two numbers, $P$ and $T$, and the equation of state produced the third. For a fixed amount of ideal gas, only two of $P$, $V$ and $T$ are independent — the equation of state uses up the third degree of freedom. That is what makes a $P$-$V$ diagram sufficient: fix a point on it and the temperature is already decided.

5. Which are state variables? $P$, $V$, $T$ and $U$ all are: each has a definite value in the state described, no matter how the gas got there. The heat supplied is not. Ask "how much heat is in this cylinder?" and the question is meaningless. Heat is energy in transit across a boundary; it exists only during a process, and the amount depends on which route was taken. The same is true of work. That asymmetry — $U$ a state function, $Q$ and $W$ path functions — decides more exam questions than any formula in the chapter.

Final Answer: $V = 5.82$ L and $U = 2909.9$ J. Pressure, volume, temperature and internal energy are state variables; the heat supplied is not.

Takeaway: Two numbers fix the state of a fixed mass of ideal gas; the third comes free from $PV = nRT$. And if a quantity only makes sense while something is happening, it is a path function, not a state variable.

Part 2: The First Law in Every Combination

Seven problems on $\Delta Q = \Delta U + \Delta W$ and nothing else. Every one of them is solved by writing the law down, putting in the two quantities you know with their signs, and reading off the third. The arithmetic is trivial; the signs are everything.

Key Point: Rearranged three ways, the first law is $\Delta U = \Delta Q - \Delta W$, $\Delta W = \Delta Q - \Delta U$ and $\Delta Q = \Delta U + \Delta W$. Learn the last one and derive the other two on the spot — that way you can never lose a minus sign by misremembering.

Example 5: Heat in, work out

A gas in a cylinder absorbs 850 J of heat from a flame and, in the same process, pushes the piston out, doing 320 J of work on the surroundings. Find the change in its internal energy and say what happens to its temperature.

Solution:

1. Assign the signs before anything else. Heat is absorbed by the system, so $\Delta Q = +850$ J. The gas expands and does work on the surroundings, so $\Delta W = +320$ J.

2. Apply the first law. $$\Delta Q = \Delta U + \Delta W \qquad \Longrightarrow \qquad \Delta U = \Delta Q - \Delta W$$ $$\Delta U = 850 - 320 = +530 \text{ J}$$

3. State every sign and what it means.

• $\Delta Q = +850$ J: positive, so 850 J of heat entered the gas.
• $\Delta W = +320$ J: positive, so the gas did 320 J of work on its surroundings — it expanded.
• $\Delta U = +530$ J: positive, so the internal energy rose. For an ideal gas $U$ depends only on temperature, so the gas got hotter.

1. The bookkeeping in words. Of the 850 J supplied, 320 J was immediately spent pushing the piston out against the outside world, and the remaining 530 J stayed behind as extra random molecular energy. Nothing was created and nothing was lost.

Final Answer: $\Delta U = +530$ J; the internal energy rises and the gas warms up.

Takeaway: Heat supplied splits into internal energy plus work done. Write $\Delta Q = \Delta U + \Delta W$, mark the signs on the two knowns, and the third falls out with no thinking at all.

Example 6: Compressed and cooled at the same time

A gas is compressed by a piston, 450 J of work being done on the gas, while 600 J of heat is simultaneously removed from it by a cooling coil. Find $\Delta U$ and state what happens to the temperature.

Solution:

1. Signs first, and this is where the marks are. Heat is removed, so $\Delta Q = -600$ J. Work is done ON the gas, and our convention takes work done by the system as positive, so $\Delta W = -450$ J.

2. Apply the law. $$\Delta U = \Delta Q - \Delta W = (-600) - (-450) = -600 + 450 = -150 \text{ J}$$

3. State every sign and what it means.

• $\Delta Q = -600$ J: negative, so 600 J of heat left the gas.
• $\Delta W = -450$ J: negative, so 450 J of work was done on the gas by the piston. The gas was compressed.
• $\Delta U = -150$ J: negative, so the internal energy fell by 150 J and the gas cooled.

1. Why the answer is smaller than either input. The two effects fight each other. Compressing the gas adds 450 J to it; the coolant removes 600 J. The coolant wins by 150 J, so the net result is a modest fall in internal energy. Had the compression been 700 J instead, $\Delta U$ would have come out at $+100$ J and the gas would have got hotter even while being cooled.

2. The trap. Writing $\Delta U = -600 - 450 = -1050$ J is the standard error, and it comes from putting the magnitude of the compression work in with a minus sign already attached and then subtracting it again. Fix the sign once, at the top, and then obey the formula literally.

Final Answer: $\Delta U = -150$ J; the gas cools slightly.

Takeaway: Double negatives are the chapter's favourite trap. $-(-450)$ is $+450$; write the substitution out in full rather than doing it in your head.

Example 7: A heater against a working system

An electric heater delivers energy to a system at a steady rate of 180 W. At the same time the system performs external work at a rate of 125 J per second. At what rate is its internal energy changing, and by how much does the internal energy change in 4.0 minutes?

Solution:

1. Recognise the form. The first law works just as well per second as it does per process. Divide every term by the time interval: $$\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$$ All three are now powers, measured in watts.

2. Signs. Energy is supplied to the system, so $\frac{\Delta Q}{\Delta t} = +180$ W. The system does work on the surroundings, so $\frac{\Delta W}{\Delta t} = +125$ W.

3. Rate of change of internal energy. $$\frac{\Delta U}{\Delta t} = 180 - 125 = +55 \text{ W}$$

4. Over 4.0 minutes. First convert: $4.0$ minutes $= 240$ s. $$\Delta U = 55 \times 240 = 13\,200 \text{ J} = 13.2 \text{ kJ}$$

5. State every sign and what it means. $\Delta Q = +43.2$ kJ over the interval (heat in), $\Delta W = +30.0$ kJ (work out, done by the system), $\Delta U = +13.2$ kJ (internal energy rising, so the system is warming). Check: $43.2 = 13.2 + 30.0$. The law closes exactly.

Final Answer: the internal energy rises at 55 W, and by $13.2$ kJ in 4.0 minutes.

Takeaway: The first law in rate form is the same law with every term divided by time. Watts in, watts out, watts stored — and the units tell you instantly whether you have divided by the time or forgotten to.

Example 8: An adiabatic route and a heated route between the same two states

A gas is taken from equilibrium state $A$ to equilibrium state $B$ along a perfectly insulated path, and $27.5$ J of work is found to be done on the gas. The gas is then returned to $A$ and taken to $B$ again along a completely different, non-insulated path, during which it absorbs a net $12.4$ calories of heat. How much work is done by the gas on this second route? (Take 1 calorie $= 4.186$ J.)

Solution:

1. Do the insulated route first, because it hands you $\Delta U$ for free. "Perfectly insulated" means $\Delta Q = 0$. Work is done on the gas, so $\Delta W = -27.5$ J. $$\Delta U = \Delta Q - \Delta W = 0 - (-27.5) = +27.5 \text{ J}$$

2. Here is the whole idea of the problem. $U$ is a state function. The change $\Delta U$ from $A$ to $B$ therefore has the same value $+27.5$ J on every path between those two states, insulated or not, fast or slow, in one step or in fifty.

3. Convert the heat on the second route into joules. Never mix calories and joules in the same equation: $$\Delta Q = 12.4 \text{ cal} \times 4.186 \text{ J/cal} = 51.906 \text{ J}$$ Positive, because the gas absorbs it.

4. Now the first law on the second route. $$\Delta W = \Delta Q - \Delta U = 51.906 - 27.5 = +24.406 \text{ J} \approx +24.4 \text{ J}$$

5. State every sign and what it means.

• Route 1: $\Delta Q = 0$ (insulated), $\Delta W = -27.5$ J (work done on the gas), $\Delta U = +27.5$ J (internal energy rises).
• Route 2: $\Delta Q = +51.9$ J (heat absorbed), $\Delta W = +24.4$ J (work done by the gas — it expands on this route), $\Delta U = +27.5$ J, identical to route 1.

1. Read the contrast. On one path the gas is squeezed and does negative work; on the other it expands and does positive work. $\Delta Q$ and $\Delta W$ are wildly different on the two routes — they are path functions. Only their difference is fixed.

Final Answer: $+24.4$ J of work is done by the gas on the second route.

Takeaway: Find $\Delta U$ on whichever path is easiest, then carry it to the path you were actually asked about. An adiabatic leg, where $\Delta U = -\Delta W$, is nearly always the easy one.

Example 9: Heat, work or a paddle wheel — three ways to add the same energy

A fixed mass of gas is to be taken from state $A$ to state $B$, for which $\Delta U = +400$ J, by three different routes.

• Route 1: the gas is heated, absorbing 700 J, while it is allowed to expand.
• Route 2: the gas is compressed, 150 J of work being done on it, while heat is also supplied.
• Route 3: the cylinder is perfectly insulated and a paddle wheel inside it is turned by a falling weight.

Find $\Delta W$ on routes 1 and 3 and $\Delta Q$ on route 2, and comment on route 3.

Solution:

1. The one fact common to all three. $U$ is a state function, so $\Delta U = +400$ J on every route. That is the anchor.

2. Route 1. $\Delta Q = +700$ J (absorbed). $$\Delta W = \Delta Q - \Delta U = 700 - 400 = +300 \text{ J}$$ Positive, so the gas does 300 J of work on the surroundings — it expands, as stated.

3. Route 2. Work is done on the gas, so $\Delta W = -150$ J. $$\Delta Q = \Delta U + \Delta W = 400 + (-150) = +250 \text{ J}$$ Positive, so 250 J of heat still has to be supplied. Less heat is needed than on route 1, because the compression is already delivering 150 J of the required 400 J.

4. Route 3. Perfect insulation means $\Delta Q = 0$. $$\Delta W = \Delta Q - \Delta U = 0 - 400 = -400 \text{ J}$$ Negative, so 400 J of work is done on the gas — by the paddle wheel, which is driven by the falling weight. No heat crosses the boundary at all; the entire internal energy rise is delivered mechanically.

5. Why route 3 is special, and worth a sentence in the exam. The paddle wheel does not push a piston; it stirs. The gas passes through states with no well-defined pressure or temperature while the stirring is going on, so route 3 cannot be drawn as a curve on a $P$-$V$ diagram at all. It is adiabatic but emphatically not quasi-static, and it is irreversible — no amount of waiting will make the gas spontaneously turn the paddle backwards and lift the weight. That does not stop the first law from applying: the first law never asks whether a process is reversible.

6. The three ledgers side by side.

Route	$\Delta Q$ (J)	$\Delta U$ (J)	$\Delta W$ (J)
1: heated, expands	$+700$	$+400$	$+300$
2: heated, compressed	$+250$	$+400$	$-150$
3: insulated, stirred	$0$	$+400$	$-400$

Every row obeys $\Delta Q = \Delta U + \Delta W$. The middle column never moves; the outer two roam freely.

Final Answer: route 1, $\Delta W = +300$ J done by the gas; route 2, $\Delta Q = +250$ J absorbed; route 3, $\Delta W = -400$ J, i.e. 400 J done on the gas by the paddle.

Takeaway: The same $\Delta U$ can be bought with pure heat, pure work, or any mixture. That is precisely what "state function" means, and it is the single most examinable sentence in this chapter.

Example 10: From joules to a temperature rise

Two moles of a diatomic ideal gas absorb 1500 J of heat and do 900 J of work on their surroundings. By how much does the temperature rise? (Take $R = 8.314$ J/(mol K).)

Solution:

1. Signs. $\Delta Q = +1500$ J (absorbed); $\Delta W = +900$ J (done by the gas, so it expanded).

2. First law for the internal energy. $$\Delta U = \Delta Q - \Delta W = 1500 - 900 = +600 \text{ J}$$ Positive, so the gas warms.

3. Convert internal energy into temperature. For an ideal gas, whatever the process, $$\Delta U = nC_v\,\Delta T$$ and for a diatomic gas $C_v = \frac{5}{2}R = \frac{5}{2}\times 8.314 = 20.785$ J/(mol K).

4. Solve. $$\Delta T = \frac{\Delta U}{nC_v} = \frac{600}{2.00 \times 20.785} = \frac{600}{41.57} = 14.43 \text{ K}$$

5. A point that catches people out. $\Delta U = nC_v\Delta T$ was derived for a constant-volume process, but it holds for every process on an ideal gas, because $U$ depends only on $T$. Here the gas certainly did not stay at constant volume — it expanded and did 900 J of work — and the formula is still correct. What is not correct here is $\Delta Q = nC_v\Delta T$; the heat depends on the path, so that would give the wrong answer.

6. And the units. A rise of $14.43$ K is a rise of $14.43$ Celsius degrees. A temperature difference has the same number on both scales; only a reading differs.

Final Answer: the temperature rises by $14.43$ K.

Takeaway: $\Delta U = nC_v\Delta T$ works for every process on an ideal gas; $\Delta Q = nC_v\Delta T$ works only at constant volume. Confusing the two is worth a lot of lost marks.

Example 11: A closed loop audited with nothing but the first law

A gas is taken round a closed cycle. On some legs it absorbs a total of 2400 J of heat; on the others it rejects a total of 1650 J. Find the net work done per cycle and the change in internal energy per cycle. If the same cycle is retraced in the opposite sense, what changes? Compute the efficiency in the first case and the coefficient of performance in the second.

Solution:

1. The defining property of a cycle. The gas returns to exactly its starting state, so every state variable returns to its starting value. In particular $$\Delta U_{\text{cycle}} = 0$$ This is true for every cycle, of every shape, in every substance.

2. Net heat, with signs. $$\Delta Q_{\text{net}} = (+2400) + (-1650) = +750 \text{ J}$$ Positive overall, so more heat entered than left.

3. Net work. With $\Delta U = 0$ the first law collapses to $\Delta Q = \Delta W$: $$\Delta W_{\text{net}} = +750 \text{ J}$$ Positive, so the gas does 750 J of net work on its surroundings each cycle. On a $P$-$V$ diagram this is the enclosed area, and the loop is traced clockwise.

4. Efficiency as an engine. The heat drawn from the hot side is $Q_1 = 2400$ J and the work delivered is $W = 750$ J: $$\eta = \frac{W}{Q_1} = \frac{750}{2400} = 0.3125 = 31.2\%$$ Check with the other form: $1 - \frac{Q_2}{Q_1} = 1 - \frac{1650}{2400} = 0.3125$. The two agree, as they must.

5. Now run the loop backwards. Every heat and every work reverses sign. The gas now absorbs 1650 J at the cold side, rejects 2400 J at the hot side, and requires $\Delta W_{\text{net}} = -750$ J — that is, 750 J of work must be done on it each cycle. The loop is traced anticlockwise. This is a refrigerator. $$\alpha = \frac{Q_2}{W} = \frac{1650}{750} = 2.20$$

6. Read the two numbers side by side, because students constantly confuse them. As an engine the device converts 31.2% of its heat intake into work. Run backwards it moves 2.20 J of heat out of the cold space for every joule of work paid in. The second number is bigger than 1 and that is not a paradox: $\alpha$ is not an efficiency, because nothing is being converted. Heat is being moved, and moving is cheap.

Final Answer: $\Delta U = 0$ and $\Delta W = +750$ J per cycle, with $\eta = 31.2\%$. Reversed, 750 J of work must be supplied per cycle and $\alpha = 2.20$.

Takeaway: Round any loop, $\Delta U = 0$, so $\Delta Q_{\text{net}} = \Delta W_{\text{net}}$ exactly. Clockwise means an engine and positive net work; anticlockwise means a refrigerator and work paid in.

Part 3: Work as the Area Under a Pressure-Volume Curve

Five problems on the single most visual idea in the chapter: since $W = \int P\,dV$, the work done by the gas is the area under its path on a $P$-$V$ diagram. Get the geometry right and half of thermodynamics becomes mensuration.

Key Point: Area is measured down to the volume axis, not to the origin, and it carries a sign. Moving right (expansion) counts the area as positive; moving left (compression) counts it as negative. A closed loop therefore encloses the net work: positive clockwise, negative anticlockwise.

Example 12: A sloping leg, then an isobaric squeeze

A gas is taken from state $D$ at $(0.020$ m$^3$, $6.0\times10^{5}$ Pa$)$ to state $E$ at $(0.050$ m$^3$, $3.0\times10^{5}$ Pa$)$ along a straight line on the $P$-$V$ diagram. From $E$ its volume is then reduced back to $0.020$ m$^3$ at constant pressure, arriving at state $F$. Find the work done by the gas on each leg and in total.

Solution:

1. Leg $D \to E$: a trapezium. The path is a straight line, so the area under it is a trapezium of parallel sides $P_D$ and $P_E$ and width $\Delta V$: $$W_{DE} = \frac{1}{2}\left(P_D + P_E\right)\left(V_E - V_D\right)$$ $$W_{DE} = \frac{1}{2}\left(6.0\times10^{5} + 3.0\times10^{5}\right)\left(0.050 - 0.020\right) = \frac{1}{2}\left(9.0\times10^{5}\right)\left(0.030\right)$$ $$W_{DE} = +13\,500 \text{ J}$$ Positive, because the volume increased: the gas expanded and did work on the surroundings.

2. Leg $E \to F$: a rectangle, traversed leftwards. Pressure is constant at $3.0\times10^{5}$ Pa: $$W_{EF} = P\left(V_F - V_E\right) = 3.0\times10^{5}\left(0.020 - 0.050\right) = 3.0\times10^{5}\times\left(-0.030\right)$$ $$W_{EF} = -9000 \text{ J}$$ Negative, because the volume decreased: 9000 J of work was done on the gas by whatever pushed the piston in.

3. Total. $$W_{\text{total}} = W_{DE} + W_{EF} = 13\,500 - 9000 = +4500 \text{ J}$$

4. Check the answer geometrically. The two areas overlap over the whole width; what survives is the triangle above the line $P = 3.0\times10^{5}$ Pa and below $DE$: $$\text{triangle} = \frac{1}{2}\times\left(6.0\times10^{5} - 3.0\times10^{5}\right)\times 0.030 = \frac{1}{2}\times 3.0\times10^{5}\times 0.030 = 4500 \text{ J}$$ The two routes to the answer agree exactly, which is the check worth doing.

5. State the signs and what they mean. $\Delta W_{DE} = +13\,500$ J (work done by the gas), $\Delta W_{EF} = -9000$ J (work done on the gas), net $+4500$ J done by the gas. Note that $D$ and $F$ have the same volume but different pressures, so this is not a closed cycle — do not be tempted to call $4500$ J an "enclosed area" of a loop.

Final Answer: $W_{DE} = +13\,500$ J, $W_{EF} = -9000$ J, total $= +4500$ J done by the gas.

Takeaway: Do each leg separately, sign it by the direction of travel, and add. Two-leg problems fall apart the moment you try to do them in a single formula.

Example 13: Kilopascals times litres are joules

A gas is taken along a straight line on a $P$-$V$ diagram whose axes are marked in kilopascals and litres, from $(2.0$ L, 200 kPa$)$ to $(8.0$ L, 500 kPa$)$. Find the work done by the gas.

Solution:

1. Check the unit product before converting anything. $$1 \text{ kPa} \times 1 \text{ L} = 10^{3} \text{ Pa} \times 10^{-3} \text{ m}^3 = 1 \text{ J}$$ So an area read straight off axes in kilopascals and litres is already in joules. This is worth memorising: it turns most graph-reading questions into mental arithmetic.

2. Area of the trapezium. $$W = \frac{1}{2}\left(P_1 + P_2\right)\left(V_2 - V_1\right) = \frac{1}{2}\left(200 + 500\right)\times\left(8.0 - 2.0\right)$$ $$W = 350 \times 6.0 = 2100 \text{ J}$$

3. Confirm it in SI, to prove the shortcut. $$W = \frac{1}{2}\left(2.0\times10^{5} + 5.0\times10^{5}\right)\times\left(6.0\times10^{-3}\right) = \frac{1}{2}\times 7.0\times10^{5}\times 6.0\times10^{-3} = 2100 \text{ J}$$ Identical, as promised.

4. Sign. $\Delta W = +2100$ J: positive, because the volume rose from 2.0 L to 8.0 L. The gas expanded and did 2100 J of work on its surroundings. Both the pressure and the volume increased, which means heat must have been supplied generously — $\Delta U$ is certainly positive too, since $PV$ and hence $T$ rose.

Final Answer: $W = +2100$ J done by the gas.

Takeaway: kPa $\times$ L $=$ J. Read the area off the graph in the units printed on the axes and you have the answer in joules without a single power of ten.

Example 14: The area is real, but the work is negative

A gas is compressed along a straight line on a $P$-$V$ diagram from $(0.060$ m$^3$, $1.0\times10^{5}$ Pa$)$ to $(0.020$ m$^3$, $4.0\times10^{5}$ Pa$)$. Find the work done by the gas, and the work done on the gas.

Solution:

1. Same trapezium formula, but watch the order of the volumes. Keep the subscripts in the order the process actually happens: start is state 1, end is state 2. $$W = \frac{1}{2}\left(P_1 + P_2\right)\left(V_2 - V_1\right)$$ $$W = \frac{1}{2}\left(1.0\times10^{5} + 4.0\times10^{5}\right)\left(0.020 - 0.060\right) = \frac{1}{2}\times 5.0\times10^{5}\times\left(-0.040\right)$$ $$W = -10\,000 \text{ J} = -10 \text{ kJ}$$

2. Read the sign properly. $\Delta W = -10$ kJ: negative, so the gas does $-10$ kJ of work on its surroundings, which is the same statement as 10 kJ of work is done ON the gas. The area on the diagram is of course a positive $10$ kJ as a geometrical area; the sign comes from the direction of travel, not from the geometry.

3. What that implies about the rest of the ledger. Both the pressure and the volume changed, and $P_2V_2 = 4.0\times10^{5}\times 0.020 = 8000$ J while $P_1V_1 = 1.0\times10^{5}\times 0.060 = 6000$ J. Since $PV \propto T$, the gas ended hotter, so $\Delta U > 0$. With $\Delta U > 0$ and $\Delta W = -10$ kJ, the first law gives $\Delta Q = \Delta U + \Delta W$, which could be either sign depending on the number of moles — but if $\Delta U$ is smaller than 10 kJ, heat must have been removed even though the gas got hotter. Compression can heat a gas while it is being cooled; there is no contradiction.

4. The habitual error. Writing "work done $= 10$ kJ" and stopping. Ten kilojoules by or on? In this chapter that is not a stylistic quibble; it flips the sign of the answer to the next part of the question.

Final Answer: $\Delta W = -10$ kJ, i.e. 10 kJ of work is done on the gas.

Takeaway: The direction of travel sets the sign; the geometry only sets the size. Rightwards is positive, leftwards is negative, always.

Example 15: How far wrong is a straight line drawn under a hyperbola?

One mole of an ideal gas at 300 K expands isothermally from $0.010$ m$^3$ to $0.025$ m$^3$. Find the work done. Then find what you would have got by drawing a straight chord between the two end states instead of the true curve, and say how big the error is. (Take $R = 8.314$ J/(mol K).)

Solution:

1. The two end pressures, from $PV = nRT$. The temperature 300 K is already absolute. $$P_1 = \frac{nRT}{V_1} = \frac{1 \times 8.314\times 300}{0.010} = 2.494\times10^{5} \text{ Pa}$$ $$P_2 = \frac{nRT}{V_2} = \frac{2494.2}{0.025} = 9.977\times10^{4} \text{ Pa}$$

2. The true area under the isotherm. Because $P = \frac{nRT}{V}$ varies along the path, the area is an integral, not a trapezium: $$W = \int_{V_1}^{V_2} P\,dV = nRT\int_{V_1}^{V_2}\frac{dV}{V} = nRT\ln\frac{V_2}{V_1}$$ $$W = 1 \times 8.314 \times 300 \times \ln\frac{0.025}{0.010} = 2494.2 \times \ln 2.5 = 2494.2 \times 0.9163$$ $$W = +2285.4 \text{ J}$$

3. The straight-chord estimate. Pretend the path were a straight line between the same two states: $$W_{\text{chord}} = \frac{1}{2}\left(P_1 + P_2\right)\left(V_2 - V_1\right) = \frac{1}{2}\left(2.494\times10^{5} + 0.998\times10^{5}\right)\left(0.015\right)$$ $$W_{\text{chord}} = \frac{1}{2}\times 3.492\times10^{5}\times 0.015 = 2618.9 \text{ J}$$

4. The error. $$\frac{2618.9 - 2285.4}{2285.4} \times 100 = 14.6\%$$ The chord overestimates by nearly fifteen per cent, and it always overestimates, because a hyperbola is convex — it sags below the chord drawn across it.

5. Signs and their meaning. $\Delta W = +2285.4$ J (positive: the gas expands and does work on the surroundings). Isothermal on an ideal gas means $\Delta T = 0$, hence $\Delta U = 0$, hence $$\Delta Q = \Delta U + \Delta W = 0 + 2285.4 = +2285.4 \text{ J}$$ positive, so exactly that much heat had to be drawn in from the reservoir to keep the temperature steady. Every joule that came in as heat left as work.

Final Answer: the true work is $+2285.4$ J; the straight-chord estimate of $2618.9$ J is 14.6% too high.

Takeaway: A curved path is never safely replaced by the straight line across it. If the pressure is not constant and not linear in $V$, the area needs an integral, and for an isotherm that integral is a logarithm.

Example 16: Two staircases between the same two corners

An ideal gas is taken from state $A$ at $(0.010$ m$^3$, $4.0\times10^{5}$ Pa$)$ to state $B$ at $(0.030$ m$^3$, $1.0\times10^{5}$ Pa$)$ by two different two-leg routes:

• Path I: expand at the constant pressure $4.0\times10^{5}$ Pa to $0.030$ m$^3$, then drop the pressure at constant volume to $1.0\times10^{5}$ Pa.
• Path II: drop the pressure at constant volume to $1.0\times10^{5}$ Pa first, then expand at that constant pressure to $0.030$ m$^3$.

Find the work done on each path, the difference, and comment.

Solution:

1. Path I. The isobaric leg does all the work; the isochoric leg does none, because $\Delta V = 0$. $$W_{\text{I}} = P_{\text{high}}\,\Delta V + 0 = 4.0\times10^{5}\times\left(0.030 - 0.010\right) = +8000 \text{ J}$$

2. Path II. Now the pressure drops first, so the expansion happens against the low pressure. $$W_{\text{II}} = 0 + P_{\text{low}}\,\Delta V = 1.0\times10^{5}\times\left(0.020\right) = +2000 \text{ J}$$

3. The difference. $$W_{\text{I}} - W_{\text{II}} = 8000 - 2000 = 6000 \text{ J}$$ And look what that equals: the area of the rectangle enclosed if you go out along path I and back along path II, $$\left(P_{\text{high}} - P_{\text{low}}\right)\left(V_B - V_A\right) = \left(3.0\times10^{5}\right)\left(0.020\right) = 6000 \text{ J}$$ They match exactly, which is the check.

4. What is the same and what is different. Both paths start at $A$ and end at $B$, so $\Delta U$ is identical on the two — $U$ is a state function. But $W$ differs by 6000 J, so by the first law $\Delta Q$ must differ by 6000 J too. Path I needs 6000 J more heat than path II.

5. Signs, stated. $\Delta W_{\text{I}} = +8000$ J and $\Delta W_{\text{II}} = +2000$ J, both positive because both paths expand overall. On path I the gas is at high pressure while it expands and so pushes harder for the same displacement; that is the whole reason it does more work.

6. The moral, in one line. Work is a path function. If a question gives you two routes between the same two states, it is almost certainly testing this, and the answer is almost always "$\Delta U$ same, $\Delta Q$ and $\Delta W$ different".

Final Answer: $W_{\text{I}} = +8000$ J, $W_{\text{II}} = +2000$ J; the difference of 6000 J is the enclosed rectangle, and $\Delta U$ is the same on both paths.

Takeaway: Expand while the pressure is high and you get more work out. That single sentence is the design principle behind every heat engine ever built.

Part 4: Molar Specific Heats and Mayer's Relation

Five problems on why a gas needs two specific heats and what connects them. The whole part rests on three lines: $\Delta U = nC_v\Delta T$ for any process, $\Delta Q_P = nC_p\Delta T$ at constant pressure, and $C_p - C_v = R$.

Key Point — the four relations you will use constantly: $$C_p - C_v = R, \qquad \gamma = \frac{C_p}{C_v}, \qquad C_v = \frac{R}{\gamma - 1}, \qquad C_p = \frac{\gamma R}{\gamma - 1}$$ With $R = 8.314$ J/(mol K), these give $C_v = \frac{3}{2}R = 12.471$ and $C_p = \frac{5}{2}R = 20.785$ for a monatomic gas ($\gamma = \frac{5}{3}$), $\frac{5}{2}R = 20.785$ and $\frac{7}{2}R = 29.099$ for a diatomic gas ($\gamma = \frac{7}{5}$), and $3R = 24.94$ and $4R = 33.26$ for a polyatomic one ($\gamma = \frac{4}{3}$), all in J/(mol K).

Use the exact fractions. Feeding the rounded decimal $1.33$ into $C_v = \frac{R}{\gamma - 1}$ gives $25.2$ instead of $24.94$ — a 1% drift that is enough to make you pick the wrong option.

Example 17: Warming nitrogen under a free piston, ledger and all

How much heat must be supplied to $3.5 \times 10^{-2}$ kg of nitrogen at room temperature to raise its temperature by 50 K at constant pressure? Split the answer into the part that becomes internal energy and the part that becomes work. (Molar mass of N$_2$ is 28 g/mol; $R = 8.314$ J/(mol K).)

Solution:

1. Moles first. $$n = \frac{m}{M} = \frac{3.5\times10^{-2}}{28\times10^{-3}} = 1.25 \text{ mol}$$

2. Identify the gas type. Nitrogen is diatomic, so at room temperature $$C_v = \frac{5}{2}R = 20.785 \text{ J/(mol K)}, \qquad C_p = \frac{7}{2}R = 29.099 \text{ J/(mol K)}$$ Note $C_p - C_v = 8.314 = R$, exactly as Mayer's relation requires.

3. The heat, at constant pressure. $$\Delta Q = nC_p\,\Delta T = 1.25 \times 29.099 \times 50 = +1818.7 \text{ J}$$

4. Where it goes. The internal energy change uses $C_v$ regardless of the process: $$\Delta U = nC_v\,\Delta T = 1.25 \times 20.785 \times 50 = +1299.1 \text{ J}$$ and the work at constant pressure is $$\Delta W = P\,\Delta V = nR\,\Delta T = 1.25 \times 8.314 \times 50 = +519.6 \text{ J}$$

5. Check the first law closes. $$\Delta U + \Delta W = 1299.1 + 519.6 = 1818.7 \text{ J} = \Delta Q \quad \checkmark$$

6. State every sign. $\Delta Q = +1818.7$ J: heat entered the gas. $\Delta U = +1299.1$ J: the internal energy rose, so the gas is hotter. $\Delta W = +519.6$ J: the gas expanded and pushed the piston out, doing work on the atmosphere. Of every 100 J supplied, about 71 J stayed as internal energy and 29 J was spent on the piston — and $\frac{R}{C_p} = \frac{8.314}{29.099} = 0.286$ confirms the 28.6%.

7. A note on $\Delta T$. The rise is "50 K", which is the same as a rise of 50 Celsius degrees. Only a temperature difference appears here, so no conversion is needed. Had the problem asked for a ratio of temperatures, kelvin would have been compulsory.

Final Answer: $\Delta Q = +1818.7$ J, of which $+1299.1$ J becomes internal energy and $+519.6$ J is work done by the gas.

Takeaway: At constant pressure, the fraction of the heat that becomes work is $\frac{R}{C_p} = \frac{\gamma-1}{\gamma}$ — about 29% for a diatomic gas and 40% for a monatomic one. It never depends on how much gas there is.

Example 18: Carbon dioxide, per mole and per kilogram

Carbon dioxide is polyatomic, with $\gamma = \frac{4}{3} \approx 1.33$ and molar mass 44 g/mol. Find its molar specific heats $C_p$ and $C_v$, and its specific heats per kilogram $c_p$ and $c_v$. Verify Mayer's relation in both sets of units.

Solution:

1. Molar values from $\gamma$ alone. Combine $C_p - C_v = R$ with $C_p = \gamma C_v$: $$\gamma C_v - C_v = R \qquad \Longrightarrow \qquad C_v = \frac{R}{\gamma - 1}$$ Use the exact value $\gamma = \frac{4}{3}$, so that $\gamma - 1 = \frac{1}{3}$: $$C_v = \frac{R}{1/3} = 3R = 3 \times 8.314 = 24.94 \text{ J/(mol K)}$$ $$C_p = \gamma C_v = \frac{4}{3} \times 3R = 4R = 4 \times 8.314 = 33.26 \text{ J/(mol K)}$$

2. Check Mayer's relation. $$C_p - C_v = 4R - 3R = R = 8.314 \text{ J/(mol K)} \quad \checkmark$$ and numerically $33.26 - 24.94 = 8.32$, which is $R$ to this rounding.

3. Convert to per-kilogram values. A molar quantity divided by the molar mass in kilograms per mole gives the per-kilogram quantity: $$c_v = \frac{C_v}{M} = \frac{3R}{0.044} = 566.9 \text{ J/(kg K)}$$ $$c_p = \frac{C_p}{M} = \frac{4R}{0.044} = 755.8 \text{ J/(kg K)}$$

4. Mayer's relation in per-kilogram form. Divide $C_p - C_v = R$ through by $M$: $$c_p - c_v = \frac{R}{M} = \frac{8.314}{0.044} = 188.95 \text{ J/(kg K)}$$ and indeed $755.8 - 566.9 = 188.9$ J/(kg K). The difference is no longer $R$ in these units — it is $\frac{R}{M}$, and it is different for every gas. That is exactly why the molar form is the one worth memorising.

5. Why the notation matters. Capital $C$ means per mole, lowercase $c$ means per kilogram. Mixing them is a guaranteed factor-of-$M$ error, and $M$ for carbon dioxide is 44, so the error is not subtle.

6. A warning about the decimal. If you put $\gamma = 1.33$ into $\frac{R}{\gamma - 1}$ you get $\frac{8.314}{0.33} = 25.19$, not $24.94$. The rounding in $\gamma$ is only 0.25%, but it sits in a denominator that is itself small, so it blows up to 1% in the answer. Whenever $\gamma - 1$ appears underneath, use the fraction, never the decimal.

Final Answer: $C_v = 3R = 24.94$ and $C_p = 4R = 33.26$ J/(mol K); $c_v = 566.9$ and $c_p = 755.8$ J/(kg K). $C_p - C_v = R$, while $c_p - c_v = \frac{R}{M} = 188.95$ J/(kg K).

Takeaway: Mayer's relation is $C_p - C_v = R$ per mole and $c_p - c_v = \frac{R}{M}$ per kilogram. Check which one the question is using by looking at the units before you write a single number.

Example 19: The price of letting the piston move

A fixed amount of gas is to be warmed through the same $\Delta T$ twice: once in a rigid sealed vessel, once under a free-moving piston at constant pressure. By what percentage does the constant-pressure heating cost more, for a monatomic, a diatomic and a polyatomic gas? Explain the pattern.

Solution:

1. Write the two heats. $$\Delta Q_V = nC_v\,\Delta T, \qquad \Delta Q_P = nC_p\,\Delta T$$

2. Form the extra fraction. $$\frac{\Delta Q_P - \Delta Q_V}{\Delta Q_V} = \frac{C_p - C_v}{C_v} = \frac{R}{C_v} = \gamma - 1$$ using $C_p - C_v = R$ and $C_v = \frac{R}{\gamma - 1}$. The answer depends only on $\gamma$, not on $n$ and not on $\Delta T$.

3. Put the three values in.

Gas type	$\gamma$	$C_v$ (J/(mol K))	$C_p$ (J/(mol K))	Extra heat needed, $\gamma - 1$
Monatomic (He, Ar)	$\frac{5}{3}$	$\frac{3}{2}R = 12.471$	$\frac{5}{2}R = 20.785$	$\frac{2}{3} = 66.7\%$
Diatomic (N$_2$, O$_2$, air)	$\frac{7}{5}$	$\frac{5}{2}R = 20.785$	$\frac{7}{2}R = 29.099$	$\frac{2}{5} = 40.0\%$
Polyatomic (CO$_2$)	$\frac{4}{3}$	$3R = 24.94$	$4R = 33.26$	$\frac{1}{3} = 33.3\%$

Each $\gamma$ is written as an exact fraction on purpose. The polyatomic row in particular is $\frac{4}{3}$, not $1.33$: the rounded decimal would give $33.0\%$ here and a $C_v$ of $25.19$, both slightly wrong.

4. Where the extra heat goes. It is not lost. In every case the extra heat is exactly $nR\,\Delta T$, which is exactly the work $P\,\Delta V$ the gas does pushing the piston out. The rigid vessel gets that work for free because nothing moves.

5. Why a monatomic gas suffers most. The extra cost is always the same absolute amount, $nR\Delta T$. But a monatomic gas has the smallest $C_v$ — its molecules can only store energy as translation, with no rotation to soak any up — so that fixed extra is the largest fraction of a small base. A polyatomic molecule has many more internal ways to store energy, so the same $nR\Delta T$ is a smaller slice of a much bigger cake.

6. The sign ledger, for the isobaric case. $\Delta Q = +nC_p\Delta T$ (heat in), $\Delta U = +nC_v\Delta T$ (internal energy up), $\Delta W = +nR\Delta T$ (positive, work done by the gas as it expands). For the rigid vessel, $\Delta W = 0$ and $\Delta Q = \Delta U$ exactly.

Final Answer: the extra cost is $\gamma - 1$, i.e. 66.7% for a monatomic gas, 40.0% for a diatomic gas and 33.3% for a polyatomic gas.

Takeaway: The gap between $C_p$ and $C_v$ is always exactly $R$, but as a percentage it is biggest for the gas with the fewest ways to store energy. That is $\gamma - 1$, and it is why $\gamma$ is largest for monatomic gases.

Example 20: Identifying a gas from two per-kilogram numbers

A laboratory measurement on an unknown gas gives $c_p = 918$ J/(kg K) and $c_v = 658$ J/(kg K). Find its molar mass, its ratio $\gamma$, and identify the gas.

Solution:

1. Use Mayer's relation in per-kilogram form. Dividing $C_p - C_v = R$ by the molar mass $M$ gives $c_p - c_v = \frac{R}{M}$, so $$M = \frac{R}{c_p - c_v} = \frac{8.314}{918 - 658} = \frac{8.314}{260} = 0.03198 \text{ kg/mol}$$ $$M = 31.98 \text{ g/mol} \approx 32 \text{ g/mol}$$

2. The ratio of specific heats. $\gamma$ is a ratio, so it does not care whether the specific heats are per mole or per kilogram — the $M$ cancels: $$\gamma = \frac{c_p}{c_v} = \frac{918}{658} = 1.395 \approx 1.40$$

3. Cross-check with the molar values. $$C_p = c_pM = 918 \times 0.03198 = 29.36 \text{ J/(mol K)}, \qquad C_v = c_vM = 21.04 \text{ J/(mol K)}$$ $$C_p - C_v = 8.31 \text{ J/(mol K)} = R \quad \checkmark$$ and $C_v \approx \frac{5}{2}R = 20.785$, which is the diatomic value.

4. Identify. A molar mass of 32 g/mol and $\gamma \approx 1.4$: this is oxygen, O$_2$, and it is diatomic. The two pieces of evidence are independent — the mass comes from the difference of the specific heats, the structure from their ratio — and they agree, which is what gives the identification its confidence.

5. The check that saves you. If the arithmetic had produced $M = 4$ g/mol together with $\gamma = 1.4$, something would be wrong: helium has $M = 4$ but is monatomic with $\gamma = 1.67$. Whenever the mass and the ratio disagree about the kind of molecule, suspect an arithmetic slip.

Final Answer: $M \approx 32$ g/mol and $\gamma \approx 1.40$; the gas is oxygen, a diatomic gas.

Takeaway: The difference of the specific heats gives you the mass; their ratio gives you the structure. One measurement, two independent answers, and each checks the other.

Example 21: The same 5 kJ, two different temperature rises

Five kilojoules of heat is supplied to 3.0 moles of a diatomic ideal gas, once in a rigid container and once under a free piston at constant pressure. Find the temperature rise in each case, and for the constant-pressure case split the heat into internal energy and work. (Take $R = 8.314$ J/(mol K).)

Solution:

1. The specific heats. Diatomic, so $C_v = \frac{5}{2}R = 20.785$ and $C_p = \frac{7}{2}R = 29.099$ J/(mol K).

2. Rigid container: constant volume. No work is possible, so all 5000 J becomes internal energy: $$\Delta T_V = \frac{\Delta Q}{nC_v} = \frac{5000}{3.0\times 20.785} = \frac{5000}{62.355} = 80.19 \text{ K}$$

3. Free piston: constant pressure. Now some of the heat leaks away as work: $$\Delta T_P = \frac{\Delta Q}{nC_p} = \frac{5000}{3.0\times 29.099} = \frac{5000}{87.297} = 57.28 \text{ K}$$

4. The ratio is not a coincidence. $$\frac{\Delta T_V}{\Delta T_P} = \frac{80.19}{57.28} = 1.400 = \gamma$$ The same heat always produces a temperature rise $\gamma$ times bigger in a rigid vessel than under a piston, for any ideal gas.

5. Split the isobaric case. $$\Delta W = nR\,\Delta T_P = 3.0\times 8.314\times 57.28 = +1428.6 \text{ J}$$ $$\Delta U = nC_v\,\Delta T_P = 3.0\times 20.785\times 57.28 = +3571.4 \text{ J}$$ $$\Delta U + \Delta W = 3571.4 + 1428.6 = 5000 \text{ J} = \Delta Q \quad \checkmark$$

6. Signs, stated. Both cases: $\Delta Q = +5000$ J, heat entering. Rigid vessel: $\Delta W = 0$ (nothing moves) and $\Delta U = +5000$ J. Free piston: $\Delta W = +1428.6$ J (gas expands, work done by it) and $\Delta U = +3571.4$ J. In both the gas gets hotter, but the rigid vessel gets hotter faster because none of the energy is being spent on the outside world.

Final Answer: $80.19$ K at constant volume and $57.28$ K at constant pressure, a ratio of exactly $\gamma = 1.40$. Of the 5000 J at constant pressure, $3571.4$ J becomes internal energy and $1428.6$ J becomes work.

Takeaway: Same heat, rigid vessel, bigger temperature rise — by exactly the factor $\gamma$. If your two answers do not sit in that ratio, one of the specific heats is wrong.