PiTheory

PiTheory JEE Physics Practice & Revision Pack · Class 11 · Chapter 11 · JEE Main

Thermodynamics

Chapter-wise practice and revision for JEE Main: concise notes on the first law, the four standard processes, P–V diagrams and cycles, and heat engines and refrigerators, concept figures, and 45 selected previous-year JEE Main questions covering the major concepts and question patterns, each with a step-by-step solution.

45 selected questions4 topics4 concept figuresPreview · first 3 of 45 questions

How to study this chapter

  1. Look at the topic's figures, then read its notes (5 to 10 minutes). They give the major equations and methods the questions use.
  2. Solve the topic's questions with a timer: 2 minutes for Easy, 3 to 4 for Medium and Hard.
  3. Check the solution. Read the Point to note line even if you were right: it is the idea you will reuse.
  4. Mark each question Got it or Revise. Come back to the Revise list after two days.

Where the marks come from

Main idea of each of the 146 previous-year JEE questions in the PiTheory bank

Processes65
Heat engines33
First law32
P–V diagrams16

Contents

  1. 1Zeroth and first law, internal energy and heatQ1–Q11
  2. 2Isothermal, adiabatic, isobaric and isochoric processesQ12–Q25
  3. 3Work done, P–V diagrams and cyclic processesQ26–Q34
  4. 4Second law, heat engines, refrigerators and Carnot cycleQ35–Q45
  5. ✓Last-day checklist
01

Topic 1 · Q1–Q11

Zeroth and first law, internal energy and heat

32 of the 146 JEE PYQs in the bank test this as their main idea

Look first

gas: ΔUQ inW by gasQ = ΔU + WΔU = nCVΔT(any process)W = ∫P dV
Heat in = rise in internal energy + work out.

Quick notes

$$Q=\Delta U+W,\qquad \Delta U=nC_V\Delta T\ (\text{ideal gas, any process}),\qquad W=\int P\,dV$$

$Q$: heat given to the system; $W$: work done by the system. $U$ is a state function (path-independent); $Q$ and $W$ depend on the path. Zeroth law defines temperature; first law is energy conservation.

Gas$C_V$$C_P$$\gamma$
Monatomic$\tfrac32R$$\tfrac52R$$\tfrac53$
Diatomic (rigid)$\tfrac52R$$\tfrac72R$$\tfrac75$
Polyatomic (rigid, non-linear)$3R$$4R$$\tfrac43$
  • $C_P-C_V=R$ per mole; per unit mass $c_P-c_V=\dfrac RM$. $C_V=\dfrac{R}{\gamma-1}$.
  • Molar heat capacity of any process: $C=\dfrac{Q}{n\Delta T}=C_V+\dfrac{W}{n\Delta T}$. If $W=fQ$: $C=\dfrac{C_V}{1-f}$.
  • Process given as $VT=K$ or similar: find $P(T)$, then $W=\int P\,dV$.
  • Phase change at constant pressure: $\Delta U=mL-P\Delta V$.
  • Extensive (scale with size): $U$, $V$, $M$. Intensive: $P$, $T$, $\rho$. Extensive/mass is intensive.
Traps. Sign convention: some books write $dQ=dU-dW$ with $W$ on the gas. Heat added need not raise temperature (isothermal, melting).

Practice: first 3 of 11 questions

Q1AIEEE 2007Easy

When a system is taken from state $i$ to state $f$ along the path $iaf$, it is found that $Q = 50$ cal and $W = 20$ cal. Along the path $ibf$, $Q = 36$ cal. $W$ along the path $ibf$ is

  1. A$14$ cal
  2. B$6$ cal
  3. C$16$ cal
  4. D$66$ cal
Answer and solution

Answer: Option B: $6$ cal

Concept. Internal energy is a state function: its change between $i$ and $f$ is the same along every path. Heat and work depend on the path, but $\Delta U=Q-W$ does not.

Step 1. $\Delta U$ from path $iaf$. $\Delta U=50-20=30$ cal.

Step 2. Path $ibf$. Same $\Delta U$, so $W=Q-\Delta U=36-30=6$ cal.

Answer. Option B.

Point to note. $\Delta U$ same on every path.

Q2AIEEE 2007Easy

If $c_P$ and $c_V$ denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then

  1. A$c_P - c_V = \dfrac{R}{28}$
  2. B$c_P - c_V = \dfrac{R}{14}$
  3. C$c_P - c_V = R$
  4. D$c_P - c_V = 28R$
Answer and solution

Answer: Option A: $c_P - c_V = \dfrac{R}{28}$

Concept. Per mole, $C_P-C_V=R$ (Mayer's relation). Specific heats per unit mass are molar values divided by the molar mass $M$, so $c_P-c_V=\dfrac RM$.

Step 1. Molar mass of nitrogen. $N_2$: $M=28$.

Answer. $c_P-c_V=\dfrac{R}{28}$ (Option A).

Point to note. Per unit mass: $c_P-c_V=R/M$.

Q3JEE Main 2024 (6 Apr, Shift 2)Easy

A total of $48\ \text{J}$ of heat is given to one mole of helium kept in a cylinder. The temperature of the helium increases by $2^\circ\text{C}$. The work done by the gas is (Given $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$)

  1. A$24.9\ \text{J}$
  2. B$72.9\ \text{J}$
  3. C$48\ \text{J}$
  4. D$23.1\ \text{J}$
Answer and solution

Answer: Option D: $23.1\ \text{J}$

Concept. First law: $Q=\Delta U+W$. For an ideal gas $\Delta U=nC_V\Delta T$ in any process. Helium is monatomic: $C_V=\tfrac32R$.

Step 1. Change in internal energy. $\Delta U=1\times\tfrac32\times8.3\times2=24.9$ J.

Step 2. Work. $W=Q-\Delta U=48-24.9=23.1$ J.

Answer. Option D.

Point to note. $W=Q-nC_V\Delta T$.

42 more questions and 3 more topics in the full chapter

  1. 2Isothermal, adiabatic, isobaric and isochoric processes14 questions with solutions
  2. 3Work done, P–V diagrams and cyclic processes9 questions with solutions
  3. 4Second law, heat engines, refrigerators and Carnot cycle11 questions with solutions
  4. ✓Last-day checklist

The complete Thermodynamics chapter is in the Class 11 Pack and the Complete Pack, as a PDF on WhatsApp. Want to see a whole chapter first? Laws of Motion and Current Electricity are free in full.

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