Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics) — page 4.
Q13 (2025)
The magnitude of heat exchanged by a system for the given cyclic process ABCA (as shown in figure {{IMG1}}) is (in SI unit)
- $10\pi$
- $5\pi$
- zero
- $40\pi$
### Related Formula
From the first law of thermodynamics for a complete cycle:
$$\Delta U = 0 \implies Q = W = \text{Area of the loop}$$
### Core Logic
The graph shows a semicircle in a $P\text{-}V$ indicator diagram. {{SOL_IMG1}}
- Pressure dimension diameter: $\Delta P = 400 - 200 = 200\ \mathrm{kPa} = 200 \times 10^{3}\ \mathrm{Pa}$
- Volume dimension diameter: $\Delta V = 400 - 200 = 200\ \mathrm{cc} = 200 \times 10^{-6}\ \mathrm{m^3}$
Radius along Pressure axis: $R_P = 100 \times 10^3\ \mathrm{Pa}$
Radius along Volume axis: $R_V = 100 \times 10^{-6}\ \mathrm{m^3}$
Area of the closed semicircular path:
$$W = \frac{1}{2} \pi R_P R_V$$
$$W = \frac{1}{2} \times \pi \times (100 \times 10^3) \times (100 \times 10^{-6})$$
$$W = \frac{10\pi}{2} = 5\pi\ \mathrm{J}$$
Since $Q = W$, the heat exchanged has a magnitude of $5\pi\ \mathrm{J}$.
### Pattern Recognition
For a cycle on an indicator chart with mismatched scales, use the elliptic area template $\pi a b$ (or $\frac{1}{2}\pi a b$ for a half-ellipse/semicircle).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q20 (2025)
An ideal gas goes from an initial state to final state. During the process, the pressure of gas increases linearly with temperature.
A. The work done by gas during the process is zero.
B. The heat added to gas is different from change in its internal energy.
C. The volume of the gas is increased.
D. The internal energy of the gas is increased.
E. The process is isochoric (constant volume process)
Choose the correct answer from the options given below :-
- A, B, C, D Only
- A, D, E Only
- E Only
- A, C Only
### Related Formula
From the Ideal Gas Law:
$PV = nRT$
According to the First Law of Thermodynamics:
$$\Delta Q = \Delta U + W$$
### Core Logic
The question states that pressure increases linearly with temperature, which means their ratio is constant :
$$P = kT \implies \frac{P}{T} = \text{constant}$$
Since $\frac{P}{T} = \frac{nR}{V}$, the volume $V$ must remain constant throughout the process. This identifies it as an isochoric process (Statement E is true).
### Step 1: Evaluate All Statements
* Statement A: True. In an isochoric process, $dV = 0 \implies W = \int P dV = 0$.
* Statement B: False. Since work is zero, the First Law simplifies to $\Delta Q = \Delta U$, meaning heat added equals the change in internal energy.
* Statement C: False. Volume is constant, so it does not increase.
* Statement D: True. As pressure increases linearly with temperature, temperature increases, which causes the internal energy of the gas to increase.
### Step 2: Final Selection
Gathering the true statements (A, D, and E) points directly to Option (2).
### Pattern Recognition
A linear $P-T$ line passing through the origin always indicates a constant volume graph. For constant volume graphs, work done is zero, which simplifies the first law calculation.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q25 (2025)
The temperature of 1 mole of an ideal monoatomic gas is increased by $50^{\circ}C$ at constant pressure. The total heat added and change in internal energy are $E_{1}$ and $E_{2}$, respectively. If $\frac{E_{1}}{E_{2}}=\frac{x}{9}$ then the value of x is
### Related Formula
For an ideal gas thermodynamic process:
* Total heat added at constant pressure (isobaric process) is :
$$E_{1} = n C_{P} \Delta T$$
* Total change in internal energy is given by :
$$E_{2} = n C_{V} \Delta T$$
The ratio of specific heat capacities is defined as:
$$\gamma = \frac{C_{P}}{C_{V}} $$
### Core Logic
Taking the ratio of the two energy expressions[cite: 179, 823]:
$$\frac{E_{1}}{E_{2}} = \frac{n C_{P} \Delta T}{n C_{V} \Delta T} = \frac{C_{P}}{C_{V}} = \gamma$$
### Step 1: Evaluating for a Monoatomic Gas
For an ideal monoatomic gas, the degrees of freedom are $f = 3$. This gives an adiabatic index of :
$$\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3} $$
Equating this value to the given ratio expression [cite: 179, 826]:
$$\frac{5}{3} = \frac{x}{9}$$
$$x = \frac{5 \times 9}{3} = 15$$
### Pattern Recognition
The ratio of heat added to the change in internal energy during an isobaric process is always equal to the adiabatic exponent $\gamma$ of the gas.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q1294 (2025)
The workdone in an adiabatic change in an ideal gas depends upon only :
- change in its pressure
- change in its specific heat
- change in its volume
- change in its temperature
### Related Formula
$$\Delta W = -\Delta U = -n C_v \Delta T$$
### Core Logic
In an adiabatic system, no heat exchange occurs ($Q=0$). By the first law of thermodynamics, $\Delta W = -\Delta U$. Since internal energy $U$ depends explicitly on temperature metrics, the total work output shifts uniquely based on temperature variation $\Delta T$.
### Chapter Mix
Class 11 Physics: Thermodynamics
Q39 (2024)
The pressure and volume of an ideal gas are related as $PV^{3/2} = K$ (Constant). The work done when the gas is taken from state A ($P_1, V_1, T_1$) to state B ($P_2, V_2, T_2$) is:
- $2(P_1V_1 - P_2V_2)$
- $2(P_2V_2 - P_1V_1)$
- $2(\sqrt{P_1}V_1 - \sqrt{P_2}V_2)$
- $2(P_2\sqrt{V_2} - P_1\sqrt{V_1})$
### Related Formula
Work done in a polytropic process $PV^x = \text{constant}$:
$$W = \frac{P_2V_2 - P_1V_1}{1 - x}$$
### Core Logic
Here, the polytropic exponent is $x = \frac{3}{2}$.
Substituting this exponent value into the generic polytropic work integral:
$$W = \frac{P_2V_2 - P_1V_1}{1 - \frac{3}{2}} = \frac{P_2V_2 - P_1V_1}{-\frac{1}{2}}$$
### Step 1: Simplify Sign Conventions
$$W = -2(P_2V_2 - P_1V_1) = 2(P_1V_1 - P_2V_2)$$
This represents the default standard convention for work executed by the gas medium.
### Pattern Recognition
Polytropic model work formula is extremely useful. If $x > 1$, the denominator $1-x$ flips the index sequence from $(2-1)$ to $(1-2)$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
