Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics) — page 2.
Q11 (2025)
A monoatomic gas having $\gamma = \frac{5}{3}$ is stored in a thermally insulated container and the gas is suddenly compressed to $\left(\frac{1}{8}\right)^{\mathrm{th}}$ of its initial volume. The ratio of final pressure and initial pressure is:
($\gamma$ is the ratio of specific heats of the gas at constant pressure and at constant volume)
- $16$
- $40$
- $32$
- $28$
### Related Formula
$$P_i V_i^{\gamma} = P_f V_f^{\gamma}$$
where,
$P_i, P_f$ = initial and final pressures
$V_i, V_f$ = initial and final volumes
$\gamma$ = adiabatic exponent
### Core Logic
Since the gas is stored in a "thermally insulated container" and is compressed "suddenly", the process is **adiabatic**.
From the adiabatic relation:
$$\frac{P_f}{P_i} = \left(\frac{V_i}{V_f}\right)^{\gamma}$$
Given:
- $V_f = \frac{1}{8} V_i \implies \frac{V_i}{V_f} = 8$
- $\gamma = \frac{5}{3}$
### Step 1: Computation
Substitute the values to find the pressure ratio:
$$\frac{P_f}{P_i} = (8)^{5/3} = \left(2^3\right)^{5/3}$$
$$\frac{P_f}{P_i} = 2^5 = 32$$
### Pattern Recognition
Sees: "suddenly compressed" or "thermally insulated container" → Adiabatic process.
Shortcut: $P V^\gamma = \text{constant}$. Since the volume goes down by $8$ times, the pressure increases by $8^{\gamma} = 8^{5/3} = 32$ times. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q377 (2025)
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).<br style=""><br style="">**Assertion** (A): With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.<br style=""><br style="">Reason (R): In isothermal process, $PV = \text{constant}$, while in adiabatic process $PV^{\gamma} = \text{constant}$. Here $\gamma$ is the ratio of specific heats, $P$ is the pressure and $V$ is the volume of the ideal gas.<br style=""><br style="">In the light of the above statements, choose the correct answer from the options given below:
- $\text{Both (A) and (R) are true but (R) is NOT the correct explanation of (A)}$
- $\text{(A) is true but (R) is false}$
- $\text{Both (A) and (R) are true and (R) is the correct explanation of (A)}$
- $\text{(A) is false but (R) is true}$
### Related Formula
$$\left(\frac{dP}{dV}\right)_{\text{isothermal}} = -\frac{P}{V}$$
$$\left(\frac{dP}{dV}\right)_{\text{adiabatic}} = -\gamma \frac{P}{V}$$
### Core Logic
The slope of an adiabatic process on a $P-V$ diagram is $\gamma$ times steeper than that of an isothermal process:
$$\left|\left(\frac{dP}{dV}\right)_{\text{adiabatic}}\right| > \left|\left(\frac{dP}{dV}\right)_{\text{isothermal}}\right|$$
{{SOL_IMG1}}
When pressure increases (compression), the volume drops. Because the adiabatic curve is steeper, the pressure rises more rapidly for a given drop in volume, or conversely, for a specified increase in pressure, the volume falls off more rapidly in the isothermal process than the adiabatic process. Hence, Assertion (A) is true.
Reason (R) states the governing equations $PV = C$ and $PV^{\gamma} = C$, which directly lead to these slope expressions via differentiation. Thus, Reason (R) is true and correctly explains Assertion (A).
### Pattern Recognition
Adiabatic curves are steeper than isothermal curves on a $P-V$ diagram because $\gamma > 1$. For any expansion or compression process, remember that slope magnitude satisfies $\text{Slope}_{\text{adi}} = \gamma \cdot \text{Slope}_{\text{iso}}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q6 (2025)
A poly-atomic molecule ($C_V = 3R, C_P = 4R$, where $R$ is gas constant) goes from phase space point $A(P_A = 10^5\mathrm{~Pa}, V_A = 4 \times 10^{-6}\mathrm{~m}^3)$ to point $B(P_B = 5 \times 10^4\mathrm{~Pa}, V_B = 6 \times 10^{-6}\mathrm{~m}^3)$ to point $C(P_C = 10^4\mathrm{~Pa}, V_C = 8 \times 10^{-6}\mathrm{~m}^3)$. $A$ to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path. The net heat absorbed per unit mole by the system is:
{{IMG1}}
- $500 \mathrm{~R}(\ln 3 + \ln 4)$
- $450 \mathrm{~R}(\ln 4 - \ln 3)$
- $500 \mathrm{~R}\ln 2$
- $400 \mathrm{~R} \ln 4$
### Related Formula
$$\Delta Q_{\text{net}} = \Delta Q_{AB} + \Delta Q_{BC}$$
$$\Delta Q_{\text{isothermal}} = nRT \ln\left(\frac{V_f}{V_i}
ight) = P_i V_i \ln\left(\frac{V_f}{V_i}
ight)$$
### Core Logic
For path $A \rightarrow B$:
Since it is given as an adiabatic path:
$$\Delta Q_{AB} = 0$$
For path $B \rightarrow C$:
Since it is given as an isothermal path, the change in internal energy $\Delta U_{BC} = 0$. From the first law of thermodynamics, heat absorbed equals work done:
$$\Delta Q_{BC} = W_{BC} = nRT_B \ln\left(\frac{V_C}{V_B}
ight)$$
Using the ideal gas state at point $B$, $nRT_B = P_B V_B$:
$$P_B V_B = (5 \times 10^4 \mathrm{~Pa}) \times (6 \times 10^{-6} \mathrm{~m}^3) = 0.3 \mathrm{~J}$$
Wait, let's express it in terms of the gas constant $R$ for a single mole ($n=1$) using temperature data directly provided in the original figure labels ($T_B = 450\mathrm{~K}$):
$$\Delta Q_{BC} = (1) \cdot R \cdot (450) \cdot \ln\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right)$$
$$\Delta Q_{BC} = 450 R \ln\left(\frac{4}{3}
ight) = 450 R (\ln 4 - \ln 3)$$
Thus, the total net heat absorbed per unit mole is:
$$\Delta Q = 0 + 450 R (\ln 4 - \ln 3) = 450 R (\ln 4 - \ln 3)$$
### Pattern Recognition
Adiabatic paths have zero heat exchange by baseline definition. The calculation boils down directly to the work done during the isothermal stage $B \rightarrow C$ matching $RT \ln(V_f/V_i)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q7 (2025)
A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine $\mathrm{E}_1$ works between 473K to 373K and engine $\mathrm{E}_2$ works between 373K to 273K. If $\eta_{12}$ , $\eta_1$ and $\eta_2$ are the efficiencies of the engines E, $\mathrm{E}_1$ and $\mathrm{E}_2$ , respectively, then
- $\eta_{12} < \eta_1 + \eta_2$
- $\eta_{12} = \eta_1\eta_2$
- $\eta_{12} = \eta_{1} + \eta_{2}$
- $\eta_{12} \geq \eta_1 + \eta_2$
### Related Formula
$$\eta = 1 - \frac{\mathrm{T}_L}{\mathrm{T}_H}$$
### Core Logic
Let's compute the efficiency parameters explicitly:
$$\eta_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$$
$$\eta_1 = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211$$
$$\eta_2 = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268$$
Evaluating the linear sum of fractional bounds:
$$\eta_1 + \eta_2 = 0.211 + 0.268 = 0.479$$
Comparing the outputs clearly demonstrates:
$$\eta_{12} < \eta_1 + \eta_2$$
### Step 1: Final Conclusion
Thus, the inequality satisfies option (1).
### Pattern Recognition
The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as $(1-\eta_{12}) = (1-\eta_1)(1-\eta_2)$, which algebraically forces $\eta_{12} = \eta_1 + \eta_2 - \eta_1\eta_2 < \eta_1 + \eta_2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q761 (2025)
A piston of mass $M$ is hung from a massless spring whose restoring force law goes as $F = -kx^3$, where $k$ is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with '$n$' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature $T$) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $L_0$ to $L_1$, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
{{IMG1}}
- $3nRT\ln \left(\frac{L_1}{L_0}\right) + 2Mg(L_1 - L_0) + \frac{k}{3} (L_1^3 - L_0^3)$
- $nRT\ln \left(\frac{L_1^2}{L_0^2}\right) + \frac{Mg}{2} (L_1 - L_0) + \frac{k}{4} (L_1^4 - L_0^4)$
- $nRT\ln \left(\frac{L_1}{L_0}\right) + Mg(L_1 - L_0) + \frac{k}{4} (L_1^4 - L_0^4)$
- $nRT\ln \left(\frac{L_1}{L_0}\right) + Mg(L_1 - L_0) + \frac{3k}{4} (L_1^4 - L_0^4)$
### Related Formula
First Law of Thermodynamics:
$$\Delta Q = \Delta U + W_{\text{by gas}}$$
Work done by an ideal gas during isothermal expansion:
$$W_{\text{gas}} = nRT\ln\left(\frac{V_1}{V_0}\right) = nRT\ln\left(\frac{L_1}{L_0}\right)$$
Conservation of Energy (Work-Energy Theorem):
Total energy delivered by the heating filament ($W_{\text{filament}}$) must equal the total work needed to lift the piston against gravity and compress the non-linear spring.
### Core Logic
Since the process is isothermal, the change in internal energy of the ideal gas is zero ($\Delta U = 0$). Hence:
$$Q = W_{\text{gas}}$$
By the Work-Energy Theorem for the piston:
$$W_{\text{gas}} + W_{\text{filament}} = \Delta U_{\text{gravity}} + \Delta U_{\text{spring}}$$
Let's evaluate each term:
- Increase in gravitational potential energy:
$$\Delta U_{\text{gravity}} = Mg(L_1 - L_0)$$
- Increase in spring potential energy:
$$U_{\text{spring}} = -\int_{L_0}^{L_1} F_{\text{restoring}}\,dx = \int_{L_0}^{L_1} kx^3\,dx = \frac{k}{4}(L_1^4 - L_0^4)$$
### Step 1: Finding Total Energy Delivered
Isolating $W_{\text{filament}}$ (the net external energy delivered to the gas system):
$$W_{\text{filament}} = W_{\text{gas}} + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$$
Since $W_{\text{gas}} = nRT\ln\left(\frac{L_1}{L_0}\right)$:
$$W_{\text{filament}} = nRT\ln\left(\frac{L_1}{L_0}\right) + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$$
### Pattern Recognition
Notice how energy conservation instantly frames this complex thermodynamics question. The heating filament's energy simply goes into three distinct stores: the isothermal work of gas expansion, raising the mass against gravity ($Mgh$), and the potential energy of the spring (integrated from $kx^3$). Keeping this total energy ledger in mind prevents tedious mathematical tangents.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics: First Law
Class 11 Physics: Work, Energy and Power: Variable Force Integration
