Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics) — page 7.
Q39 (2026)
One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature $27^{\circ}$ C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^{\circ}$ C doing the same amount of work W, then its final temperature will be (close to) ____ ${}^{\circ}$ C.
- $-189$
- $-56$
- $-30$
- $-117$
### Related Formula
$$W_{\text{isothermal}} = nRT \ln \left(\frac{V_2}{V_1}\right)$$
$$W_{\text{adiabatic}} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$
### Core Logic
For a diatomic gas, $\gamma = 1.4$.
Given initial temperature $T = 27^{\circ}\mathrm{C} = 300 \, \mathrm{K}$.
Since $W_{\text{isothermal}} = W_{\text{adiabatic}}$, we equate the two expressions.
### Step 1: Compute Work Done in Isothermal Expansion
$$W_{\text{isothermal}} = (1) \cdot R \cdot 300 \cdot \ln(2)$$
$$W_{\text{isothermal}} = 300R(0.693) \quad \text{--- (1)}$$
### Step 2: Equate and Solve for Final Temperature
$$W_{\text{adiabatic}} = \frac{nR(T_1 - T_2)}{\gamma - 1}$$
Equating (1) to adiabatic work:
$$\frac{(1)R(300 - T_f)}{1.4 - 1} = 300R(0.693)$$
$$\frac{300 - T_f}{0.4} = 300(0.693)$$
$$300 - T_f = 0.4 \times 300 \times 0.693$$
$$300 - T_f = 120 \times 0.693 = 83.16$$
$$T_f = 300 - 83.16 = 216.84 \, \mathrm{K}$$
### Step 3: Convert to Celsius
$$T_f (\text{in } ^{\circ}\mathrm{C}) = 216.84 - 273.15 = -56.31^{\circ}\mathrm{C}$$
Closest option is $-56$.
### Pattern Recognition
Equating the work of two different processes links the log expansion to the temperature drop. Recognize diatomic gas instantly means $\gamma = 1.4$, providing the $0.4$ divisor.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q29 (2026)
Density of water at $4^{\circ}$ C and $20^{\circ}$ C are $1000 \, kg/m^{3}$ and $998 \, kg/m^{3}$ respectively. The increase in internal energy of 4 kg water when it is heated from $4^{\circ}$ C to $20^{\circ}$ C is ____J.
(Specific heat capacity of water = $4.2 \, J/g^{\circ}C$ and 1 atmospheric pressure = $10^{5}$ Pa)
- $315826.2$
- $234699.2$
- $258700.8$
- $268799.2$
### Related Formula
$$\Delta Q = mc\Delta T$$
$W = P \Delta V$
$$\Delta U = \Delta Q - W$$
### Core Logic
First, calculate the heat supplied $\Delta Q$:
$$Q = mS\Delta T = 4 \times 4200 \times 16 = 268800 \text{ J}$$
Next, calculate the work done $W$ against atmospheric pressure due to volume change:
$W = P \Delta V$
$$\Delta V = \left( \frac{m}{\rho_f} - \frac{m}{\rho_i} \right) = 4 \left[ \frac{1}{998} - \frac{1}{1000} \right]$$
Given $P = 10^5 \text{ Pa}$:
$$W = 10^5 \times 4 \times \left[ \frac{1000 - 998}{998 \times 1000} \right] = \frac{8 \times 10^5}{998 \times 10^3} \approx 0.8 \text{ J}$$
### Step 1: Internal Energy Calculation
Using the First Law of Thermodynamics:
$$\Delta U = Q - W$$
$$\Delta U = 268800 - 0.8 = 268799.2 \text{ J}$$
### Pattern Recognition
When heating solids or liquids, the expansion is so small that $\Delta V$ is negligible, meaning $W \approx 0$, making $\Delta U \approx \Delta Q$. Subtracting the exact $0.8\text{J}$ provides the highly precise answer required here.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Thermal Properties of Matter
Q43 (2026)
10 mole of an ideal gas is undergoing the process shown in the figure. The heat involved in the process from $\mathrm{P}_{1}$ to $\mathrm{P}_{2}$ is $\alpha$ Joule ( $\mathrm{P}_{1} = 21.7\mathrm{Pa}$ and $\mathrm{P}_{2} = 30\mathrm{Pa}$ , $\mathrm{C}_{\mathrm{v}} = 21\mathrm{J / K.mol}$ , $R = 8.3\mathrm{J / mol.K}$ ). The value of $\alpha$ is ____.
{{IMG1}}
- $24$
- $15$
- $21$
- $28$
### Related Formula
$$\Delta Q = n C_v \Delta T$$
From Ideal Gas Law for Isochoric Process:
$$n R \Delta T = \Delta P \cdot V$$
### Core Logic
From the P-V graph, the process from $P_1$ to $P_2$ takes place at a constant volume $V = 1 \text{ m}^3$.
### Step 1: Calculate Heat Involved
$$\Delta Q = n C_v \Delta T$$
Multiply and divide by $R$:
$$\Delta Q = \frac{C_v}{R} \cdot (nR\Delta T) = \frac{C_v}{R} \cdot (P_2 - P_1)V$$
### Step 2: Substitution
Given $C_v = 21 \text{ J/K\cdot mol}$, $R = 8.3 \text{ J/mol\cdot K}$, $P_2 = 30$, $P_1 = 21.7$, and $V = 1$:
$$\Delta Q = \frac{21}{8.3} \times (30 - 21.7) \times 1$$
$$\Delta Q = \frac{21}{8.3} \times 8.3 = 21 \text{ J}$$
### Pattern Recognition
For constant volume processes, $\Delta Q$ strictly scales with $\Delta P$. Re-writing $n \Delta T$ as $V \Delta P / R$ circumvents calculating absolute temperatures entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q46 (2026)
When 300 J of heat given to an ideal gas with $C_{p}=\frac{7}{2}R$ its temperature raises from 20 °C to 50 °C keeping its volume constant. The mass of the gas is (approximately) ____ g. (R = 8.314 J/mol.K).
### Related Formula
$C_v = C_P - R$
$$\Delta Q = n C_V \Delta T$$
### Core Logic
$$C_v = \frac{7}{2}R - R = \frac{5}{2}R$$
For an isochoric process (constant volume), the heat given is strictly equal to internal energy change:
$$300 = n \times \frac{5}{2} \times 8.314 \times (50 - 20)$$
### Step 1: Find moles (n)
$$300 = n \times \frac{5}{2} \times 8.314 \times 30$$
$$300 = n \times 75 \times 8.314$$
$$n = 0.48 \text{ moles}$$
$$\frac{m}{M} = 0.48$$
### Step 2: Evaluating the output
Our Ans. (Bonus)
We cannot definitively find the mass ($m$) in grams because the molar mass ($M$) of the specific ideal gas is not given in the problem statement.
(Note: NTA official answer is 481. This implies assuming $M \approx 1000$ or an error in problem specification).
### Pattern Recognition
Always extract $C_v$ from $C_p$ using Mayer's relation for constant volume heat transfers. If an exam sets a numerical without giving molar mass, it relies on partial contextual hints or is marked as bonus.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q37 (2026)
In the following $p - V$ diagram the equation of state along the curved path is given by $(V - 2)^2 = 4ap$ where $a$ is a constant. The total work done in the closed path is
{{IMG1}}
- $-\frac{1}{a}$
- $+\frac{1}{3a}$
- $\frac{1}{2a}$
- $-\frac{1}{3a}$
### Related Formula
$$W = \int p \, dV$$
### Core Logic
The total work done in a cyclic process is the enclosed area. The cycle is clockwise, but wait—the arrows in the diagram point anti-clockwise (A to B to C and back to A via horizontal top line). Anti-clockwise cycle means negative work. The enclosed area is bounded by a parabola.
### Step 1: Identifying the Parabola Parameters
Equation of curve: $(V - 2)^2 = 4ap$
This is an upward-opening parabola with vertex at $V = 2$, $p = 0$ (assuming $a>0$).
Let's check points A ($V=1$) and C ($V=3$).
At $V=1$: $(1 - 2)^2 = 4ap \Rightarrow 1 = 4ap_0 \Rightarrow p_0 = \frac{1}{4a}$
At $V=3$: $(3 - 2)^2 = 4ap \Rightarrow 1 = 4ap_0 \Rightarrow p_0 = \frac{1}{4a}$
So the top horizontal line is at $p_0 = \frac{1}{4a}$.
### Step 2: Area Calculation
The enclosed shape is bounded by the line $p = p_0$ and the parabola $p = \frac{(V - 2)^2}{4a}$.
The area of a parabolic segment (bounded by a chord) is $\frac{2}{3} \times \text{Area of enclosing rectangle}$.
Rectangle width = $\Delta V = 3 - 1 = 2$.
Rectangle height = $p_0 - 0 = p_0$.
Area = $\frac{2}{3} \times (p_0) \times (2) = \frac{4p_0}{3}$.
### Step 3: Substitute and Determine Sign
Substitute $p_0 = \frac{1}{4a}$:
$$|W| = \frac{4}{3} \left(\frac{1}{4a}\right) = \frac{1}{3a}$$
The cycle is counter-clockwise, so the net work done by the gas is negative.
$$W_{\text{gas}} = -\frac{1}{3a}$$
### Pattern Recognition
For parabolic curves $y = kx^2$, the area inside the 'bowl' up to a chord is $2/3$ of the bounding rectangle. Direction of arrows dictates the sign: clockwise = positive, anti-clockwise = negative.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
