Thermodynamics — Easy Practice (Set 1)

Quick Recap — Thermodynamics Basics

• Zeroth law: two bodies in thermal equilibrium have the same temperature.
• First law: $Q = \Delta U + W$, where $W$ is the work done by the gas and $\Delta U = nC_v\Delta T$.
• The four processes: isothermal ($T$ constant), isobaric ($P$ constant), isochoric ($V$ constant), adiabatic ($Q = 0$).
• Work done by a gas equals the area under its $P$-$V$ curve; isobaric $W = P\Delta V$, isochoric $W = 0$.
• Specific heats: $C_p - C_v = R$ (Mayer's relation), $\gamma = C_p/C_v$; the internal energy of an ideal gas depends only on $T$.

Beyond-NCERT JEE Formulae

A rapid-reference sheet of the high-yield thermodynamics results JEE Main leans on -- each carries a one-line when to use and a [JEE Tip] for the trap that catches most students. (The first law $Q=\Delta U+W$ and the four basic processes are assumed from the earlier sections; throughout, $W$ is the work done by the gas, $T_1$ is the hotter reservoir and $T_2$ the colder.)

1. Work done by the gas, process by process

• Isothermal ($T$ fixed): $W=nRT\ln\dfrac{V_2}{V_1}=nRT\ln\dfrac{P_1}{P_2}$
• Adiabatic ($Q=0$): $W=\dfrac{P_1V_1-P_2V_2}{\gamma-1}=\dfrac{nR(T_1-T_2)}{\gamma-1}$
• Isobaric ($P$ fixed): $W=P\,\Delta V=nR\,\Delta T$
• Isochoric ($V$ fixed): $W=0$

When to use: read off the row whose named quantity is held fixed; each result is just $W=\int P\,dV$ evaluated for that constraint.

[JEE Tip] In an adiabatic expansion $T_2<T_1$, so the work comes out positive and the gas cools; in compression it is negative. Never carry the isothermal $\ln$ formula onto an adiabatic leg -- there the temperature is not constant.

2. Adiabatic (reversible) relations

$$PV^{\gamma}=\text{const},\qquad TV^{\gamma-1}=\text{const},\qquad T^{\gamma}P^{1-\gamma}=\text{const}$$

so that $\dfrac{T_2}{T_1}=\left(\dfrac{V_1}{V_2}\right)^{\gamma-1}=\left(\dfrac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$.

When to use: any reversible $Q=0$ step in which two of $P$, $V$, $T$ change together.

[JEE Tip] Through a given point an adiabat is steeper than an isotherm: $\left(\dfrac{dP}{dV}\right)_{\text{adia}}=\gamma\left(\dfrac{dP}{dV}\right)_{\text{iso}}$. The adiabatic bulk modulus is $\gamma P$ against the isothermal $P$ -- which is exactly why the speed of sound is $v=\sqrt{\dfrac{\gamma RT}{M}}$ (Laplace), not Newton's $\sqrt{P/\rho}$.

3. Specific heats and degrees of freedom

Mayer's relation $C_P-C_V=R$ with $\gamma=\dfrac{C_P}{C_V}$. From the active degrees of freedom $f$,

$$C_V=\dfrac{f}{2}R,\qquad C_P=\left(\dfrac{f}{2}+1\right)R,\qquad \gamma=1+\dfrac{2}{f}$$

When to use: to move between $f$, $C_V$, $C_P$ and $\gamma$ -- monatomic $f=3$ ($\gamma=\tfrac53$), diatomic $f=5$ ($\gamma=\tfrac75$), nonlinear polyatomic $f=6$ ($\gamma=\tfrac43$).

[JEE Tip] Internal energy always changes as $\Delta U=nC_V\,\Delta T$ -- in every process, not only at constant volume, because $U$ depends on temperature alone.

4. Mixture of ideal gases

$$C_{V,\text{mix}}=\dfrac{n_1C_{V1}+n_2C_{V2}}{n_1+n_2},\qquad \gamma_{\text{mix}}=\dfrac{C_{V,\text{mix}}+R}{C_{V,\text{mix}}}$$

When to use: whenever two gases share one container -- build the mole-weighted $C_{V,\text{mix}}$ first, then read off $\gamma_{\text{mix}}$ from Mayer's relation.

[JEE Tip] Equivalent one-line shortcut: $\dfrac{n_1+n_2}{\gamma_{\text{mix}}-1}=\dfrac{n_1}{\gamma_1-1}+\dfrac{n_2}{\gamma_2-1}$. The mixture $\gamma$ always lands between the two component values, never outside them.

5. Polytropic process $PV^{x}=$ const

Molar heat capacity and work of a general $PV^{x}=$ const path:

$$C=C_V+\dfrac{R}{1-x}=\dfrac{R}{\gamma-1}+\dfrac{R}{1-x},\qquad W=\dfrac{nR(T_1-T_2)}{x-1}$$

When to use: any process that obeys $PV^{x}=$ const. The limits check it: $x=0$ isobaric ($C=C_P$), $x=1$ isothermal ($C\to\infty$), $x=\gamma$ adiabatic ($C=0$), $x\to\infty$ isochoric ($C=C_V$).

[JEE Tip] For $1<x<\gamma$ the molar heat capacity $C$ is negative: heat is being removed and yet the gas warms up, because its own compression work outruns the heat loss.

6. Heat engines, refrigerators and heat pumps

• Engine: $\eta=\dfrac{W}{Q_1}=1-\dfrac{Q_2}{Q_1}$, with the Carnot ceiling $\eta=1-\dfrac{T_2}{T_1}$.
• Refrigerator: $\text{COP}_{\text{ref}}=\dfrac{Q_2}{W}=\dfrac{T_2}{T_1-T_2}$.
• Heat pump: $\text{COP}_{\text{hp}}=\dfrac{Q_1}{W}=\dfrac{T_1}{T_1-T_2}$.

When to use: the temperature forms are the reversible (best-case) bounds; the heat forms hold for any real device through $Q_1=Q_2+W$.

[JEE Tip] Work only in kelvin. The refrigerator numerator carries the cold $T_2$, the heat pump the hot $T_1$, and the two are tied together by $\text{COP}_{\text{hp}}=\text{COP}_{\text{ref}}+1=\dfrac{1}{\eta_{\text{Carnot}}}$.

7. Cyclic processes and entropy

Around a closed cycle $\Delta U=0$, so $Q_{\text{net}}=W_{\text{net}}=$ the area enclosed on the $P$-$V$ diagram (positive when traversed clockwise). Between any two states of an ideal gas,

$$\Delta S=nC_V\ln\dfrac{T_2}{T_1}+nR\ln\dfrac{V_2}{V_1}$$

which for an isothermal step collapses to $\Delta S=nR\ln\dfrac{V_2}{V_1}$.

When to use: cycles -- sum the first law leg by leg; entropy -- integrate along any convenient reversible path, since $S$ is a state function.

[JEE Tip] A free (Joule) expansion into vacuum has $W=0$, $Q=0$ and $\Delta U=0$, so the temperature is unchanged -- yet $\Delta S=nR\ln\dfrac{V_2}{V_1}>0$. It is the textbook case of a process that is isothermal and irreversible at once.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Adiabatic compression: final temperature and work. Two moles of a diatomic gas ($\gamma=1.4$) at $300$ K are compressed adiabatically until the volume falls to one thirty-second of its initial value. Take $R=8.31$ J/(mol K). Find the final temperature and the work done on the gas.

Solution. Use $TV^{\gamma-1}=$ const: $T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma-1}=300\times 32^{0.4}$. Since $32=2^5$, we get $32^{0.4}=2^{2}=4$, so $T_2=300\times4=1200$ K. The work done by the gas is $W=\dfrac{nR(T_1-T_2)}{\gamma-1}=\dfrac{2(8.31)(300-1200)}{0.4}=\dfrac{-14958}{0.4}=-37395$ J. It is negative, so the work done on the gas is $+37395\approx37.4$ kJ, and since $\Delta U=-W$ the internal energy climbs by the same $37.4$ kJ.

Example 2 — Isothermal expansion: work and heat. Three moles of an ideal gas at $350$ K expand isothermally and reversibly to four times the initial volume. Take $R=8.31$ J/(mol K) and $\ln 4=1.386$. Find the work done and the heat absorbed.

Solution. Isothermal work is $W=nRT\ln\dfrac{V_2}{V_1}$. First $nRT=3\times8.31\times350=8725.5$ J, so $W=8725.5\times1.386\approx12094$ J, that is about $12.1$ kJ. The temperature is fixed, so $\Delta U=0$ and the first law gives $Q=W$: the gas draws roughly $12.1$ kJ of heat from the reservoir and returns every joule as work.

Example 3 — From the ratio $\gamma$ to $C_V$, $C_P$ and the heat at constant pressure. A gas is measured to have $\gamma=1.4$. Taking $R=8.31$ J/(mol K), find its degrees of freedom, $C_V$ and $C_P$, and the heat needed to warm $2$ mol of it through $50$ K at constant pressure.

Solution. From $\gamma=1+\dfrac{2}{f}$ we get $f=\dfrac{2}{\gamma-1}=\dfrac{2}{0.4}=5$, a diatomic gas. Then $C_V=\dfrac{f}{2}R=\tfrac52(8.31)=20.78$ J/(mol K) and $C_P=C_V+R=29.09$ J/(mol K). At constant pressure the heat is $Q=nC_P\,\Delta T=2(29.09)(50)=2909$ J. Of this, $nR\,\Delta T=2(8.31)(50)=831$ J leaves as expansion work and the remaining $2078$ J is stored as internal energy.

Example 4 — Gas mixture: $C_V$, $\gamma$ and adiabatic heating. A vessel holds $2$ mol of a monatomic gas ($C_V=\tfrac32R$) together with $3$ mol of a diatomic gas ($C_V=\tfrac52R$) at $400$ K. The mixture is compressed adiabatically to half its volume. Find $\gamma_{\text{mix}}$ and the final temperature.

Solution. Mole-weighted, $C_{V,\text{mix}}=\dfrac{2\left(\tfrac32R\right)+3\left(\tfrac52R\right)}{5}=\dfrac{3R+7.5R}{5}=2.1R$. Then $\gamma_{\text{mix}}=\dfrac{C_{V,\text{mix}}+R}{C_{V,\text{mix}}}=\dfrac{3.1R}{2.1R}=1.476$, so $\gamma_{\text{mix}}-1=0.476$. Adiabatic heating gives $T_2=T_1\left(\dfrac{V_1}{V_2}\right)^{\gamma_{\text{mix}}-1}=400\times2^{0.476}=400\times1.391=556$ K. Taking either pure gas ($\gamma=1.67$ or $1.4$) would put the exponent wrong -- the mixture value has to be built first.

Example 5 — Carnot engine: efficiency, work and rejected heat. A Carnot engine runs between a source at $227\,^\circ$C and a sink at $27\,^\circ$C, absorbing $1200$ J from the source each cycle. Find the efficiency, the work delivered and the heat rejected.

Solution. Convert to kelvin first: $T_1=227+273=500$ K and $T_2=27+273=300$ K. The Carnot efficiency is $\eta=1-\dfrac{T_2}{T_1}=1-\dfrac{300}{500}=0.40$, i.e. $40\%$. The work per cycle is $W=\eta Q_1=0.40\times1200=480$ J, and the heat rejected is $Q_2=Q_1-W=1200-480=720$ J. Had the Celsius values been used directly the efficiency would have come out wildly too large -- the formula holds only on the absolute scale.

Example 6 — Refrigerator and heat-pump COP. A Carnot refrigerator holds a chamber at $260$ K while the room sits at $300$ K, extracting $520$ J from the chamber per cycle. Find the coefficient of performance, the work input, the heat dumped into the room, and the COP if the same machine ran as a heat pump.

Solution. As a refrigerator, $\text{COP}_{\text{ref}}=\dfrac{T_2}{T_1-T_2}=\dfrac{260}{300-260}=\dfrac{260}{40}=6.5$. The work is $W=\dfrac{Q_2}{\text{COP}_{\text{ref}}}=\dfrac{520}{6.5}=80$ J, and the heat rejected to the room is $Q_1=Q_2+W=520+80=600$ J. Run instead as a heat pump, $\text{COP}_{\text{hp}}=\dfrac{T_1}{T_1-T_2}=\dfrac{300}{40}=7.5$, which is exactly $\text{COP}_{\text{ref}}+1$ and equals $\dfrac{Q_1}{W}=\dfrac{600}{80}$.

Example 7 — Polytropic process: a negative molar heat capacity. One mole of a monatomic gas ($C_V=\tfrac32R$) is taken along $PV^{1.2}=$ const while its temperature falls from $400$ K to $350$ K. Take $R=8.31$ J/(mol K). Find the molar heat capacity of the process and the work done, then confirm the first law.

Solution. With $x=1.2$, $C=C_V+\dfrac{R}{1-x}=\tfrac32R+\dfrac{R}{-0.2}=1.5R-5R=-3.5R=-29.1$ J/(mol K), which is negative. The work done by the gas is $W=\dfrac{nR(T_1-T_2)}{x-1}=\dfrac{(8.31)(50)}{0.2}=2077.5$ J, positive because the gas expands as it cools. Check the first law: $\Delta U=nC_V\,\Delta T=1.5(8.31)(-50)=-623.25$ J and $Q=nC\,\Delta T=(-29.1)(-50)=1454.25$ J, so $\Delta U+W=-623.25+2077.5=1454.25$ J, exactly the heat found above. The gas absorbs heat and yet cools -- the signature of a negative-$C$ polytrope.

Example 8 — Cyclic process: net work from the enclosed area. A gas is carried clockwise once around a rectangle on the $P$-$V$ diagram, between pressures $1\times10^{5}$ Pa and $3\times10^{5}$ Pa and volumes $2\times10^{-3}$ m$^3$ and $6\times10^{-3}$ m$^3$. Find the net work and the net heat per cycle.

Solution. Around a closed loop $\Delta U=0$, so the net work equals the area enclosed. The sides are $\Delta P=(3-1)\times10^{5}=2\times10^{5}$ Pa and $\Delta V=(6-2)\times10^{-3}=4\times10^{-3}$ m$^3$, giving $W_{\text{net}}=\Delta P\,\Delta V=(2\times10^{5})(4\times10^{-3})=800$ J, positive because the loop is clockwise. With $\Delta U=0$ the first law forces $Q_{\text{net}}=W_{\text{net}}=800$ J: over the cycle the gas takes in $800$ J of heat and delivers all of it as work. Traversed anticlockwise, the same rectangle would give $-800$ J.