JEE Main Pattern Practice Questions

Close the Notes. Start the Clock.

Sections 1 to 13 taught you this chapter — systems and state variables, the zeroth law, heat and work against internal energy, the first law, $C_p$ and $C_v$ and Mayer's relation, the $P$-$V$ diagram and work as the area under it, the four processes, the second law in both its statements, engines, refrigerators and heat pumps, the Carnot ceiling, a full set of worked problems and an advanced toolkit on top.

This section asks one different question: can you use any of it with a timer running?

There is no new theory below. There are 30 single-correct questions built to the exam pattern, and a marking scheme designed to punish the three habits this chapter rewards most cruelly — putting a Celsius number where a ratio demands kelvin, letting the two sign conventions get mixed inside one solution, and reading "$\Delta Q = 0$" as though it meant "$\Delta T = 0$".

The rules of engagement

Key Point: This is not a reading exercise. Blank sheet, pen, timer. Attempt all 30 questions in one unbroken sitting, and do not open a single explanation until the last answer is written.

The Sign Convention

Half the traps below turn on it.

Key Point: $\Delta Q$ is positive when heat is added TO the system. $\Delta W$ is positive when work is done BY the system. The first law is therefore $$\Delta Q = \Delta U + \Delta W$$ Chemistry books write $\Delta U = Q + W$ with $W$ the work done ON the system — same physics, one sign moved. Every question and every answer below uses the form above. If a question mentions work, it says by or on; if your answer has a sign, say what that sign means.

The constants sheet

Every question that needs a number uses these and no others. Copy them to the top of your sheet before you start.

The constants used throughout this drill. Nothing else is needed.

A note on symbols, so nothing is ambiguous under time pressure. $T$ is always an absolute temperature in kelvin and $t$ or $t_C$ a Celsius one. $T_1$ is the source (hot) and $T_2$ the sink (cold); $Q_1$ is absorbed from the source and $Q_2$ rejected to the sink. $n$ is the number of moles. $C_p$ and $C_v$ are molar specific heats in J/(mol K), lower-case $c_p$ and $c_v$ are per kilogram, and $\gamma = \frac{C_p}{C_v}$. $\eta$ is efficiency and $\alpha$ is a coefficient of performance — never call $\alpha$ an efficiency. $U$ is a state function; $Q$ and $W$ are path functions.

What this set covers

That spread mirrors how the paper actually samples this chapter. The first law and the four processes together are 20 of the 30, because those two blocks carry the most multi-step questions and therefore the most marks, and because a slip in either ruins everything downstream.

The difficulty mix is roughly 25% easy, 45% medium and 30% hard. A handful will feel brutal. They are meant to.

[Exam Tip] That $-1$ changes the arithmetic of guessing. A blind guess among four options returns $\frac{4}{4} - \frac{3}{4} = +0.25$ marks on average — barely worth the minute it costs. A question narrowed to two options returns $\frac{4-1}{2} = +1.50$ marks on average, six times as much. Narrow first, then commit. Leave blank only what you could not narrow at all.

[Exam Tip] Before you start, write five lines at the top of your sheet: kelvin or Celsius?, by the gas or on the gas?, is this $C_p$ or $C_v$?, did I check $\Delta Q = \Delta U + \Delta W$?, is my efficiency below $1 - \frac{T_{\min}}{T_{\max}}$? Those five questions catch the overwhelming majority of the marks lost in this chapter.

Scoring Yourself Honestly

Mark your sheet with the real scheme — $+4$, $-1$, $0$ — and total it. No half marks for "I knew that one really". The number you get is the number that matters.

The bands

Read your own answer sheet

Before you touch a single explanation, sort your mistakes into three piles. This is the most valuable ten minutes in the whole section.

1. Method errors — you put a Celsius temperature into a ratio, an efficiency or a gas-law step. You mixed the two sign conventions inside one solution. You read $\Delta Q = 0$ as $\Delta T = 0$, or applied $PV^{\gamma} =$ constant to a free expansion. You put the net heat in the denominator of an efficiency instead of the heat absorbed. These are the expensive ones, because the whole solution is wrong from line one.
2. Execution errors — right method, wrong arithmetic. The classic four in this chapter: $C_v$ used where $C_p$ was wanted; litres left unconverted inside $P\,\Delta V$; a chord read off a curve where the area had to be integrated; and $\gamma - 1$ written as $1 - \gamma$.
3. Reading errors — the question asked for the work done on the gas, not by it; for the heat rejected, not absorbed; for the coefficient of performance of the heat pump, not the refrigerator; for the temperature of the remaining gas; for the maximum temperature on a path, not the temperature at its end.

Key Point: In this chapter pile 3 is unusually fat, because so many quantities come in near-identical pairs: kelvin against Celsius, by against on, absorbed against rejected, $C_p$ against $C_v$, efficiency against coefficient of performance, source against sink. Underline the quantity the question actually wants before you start solving.

The eight habits this set is drilling

• Write K next to every absolute temperature. Ratios, efficiencies and gas-law steps all demand kelvin. Only a difference $\Delta T$ may be quoted in either scale, and then the number is the same.
• Never write a bare number for work. Write "$\Delta W = -450$ J, so 450 J of work is done ON the gas". The sentence is its own check.
• $\Delta U = nC_v\Delta T$ for an ideal gas, whatever the process. The $C_v$ is not a claim that the volume was constant.
• $\Delta U = \dfrac{P_2V_2-P_1V_1}{\gamma-1}$ gets you the internal energy change from the two corners of a path, with no moles and no temperatures.
• Work is the area under the path. A chord is exact only for a straight line; a curve has to be integrated.
• Round any closed cycle $\Delta U = 0$, so $\Delta Q = \Delta W$, and the net work is the enclosed area — positive clockwise, negative anticlockwise.
• In $\eta = \frac{W}{Q_1}$, the $Q_1$ is the heat ABSORBED, summed over only the legs where heat went in.
• Check every efficiency against $1 - \dfrac{T_{\min}}{T_{\max}}$. An engine that beats it is an arithmetic mistake, not a discovery.

[Exam Tip] Every explanation below is a full step-by-step solution, so this set doubles as revision. Read the explanation even for the questions you got right — several of these have a two-line route and a two-page route, and it is the two-line route you will need in the hall.