Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics) — page 5.
Q40 (2024)
$0.08\text{ kg}$ of air is heated at constant volume through $5^{\circ}\text{C}$. The specific heat of air at constant volume is $0.17\text{ kcal/kg}^{\circ}\text{C}$ and $J = 4.18\text{ joule/cal}$. The change in its internal energy is approximately:
- $318\text{ J}$
- $298\text{ J}$
- $284\text{ J}$
- $142\text{ J}$
### Related Formula
$$Q = \Delta U + W$$
Since the volume is constant, work done $W = 0$, leading to:
$$\Delta U = Q = m c_v \Delta T$$
### Core Logic
Given values:
$m = 0.08\text{ kg}$
$c_v = 0.17\text{ kcal/kg}^{\circ}\text{C} = 0.17 \times 10^3\text{ cal/kg}^{\circ}\text{C}$
$\Delta T = 5^{\circ}\text{C}$
$J = 4.18\text{ J/cal}$
### Step 1: Metric conversion and computing value
$$\Delta U = 0.08 \times (0.17 \times 10^3) \times 5 \times 4.18$$
$$\Delta U = 0.08 \times 170 \times 5 \times 4.18$$
$$\Delta U = 68 \times 4.18 \simeq 284.24\text{ J} \approx 284\text{ J}$$
### Pattern Recognition
Isochoric processes channel total thermal input strictly into core state configurations, converting raw metric calories directly to standard mechanical Joules via mechanical equivalent factors ($J$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q1636 (2024)
A thermodynamic system is taken from an original state A to an intermediate state B by a linear process as shown in the figure. It's volume is then reduced to the original value from B to C by an isobaric process. The total work done by the gas from A to B and B to C would be:
{{IMG1}}
- $33800 \mathrm{~J}$
- $2200 \mathrm{~J}$
- $600 \mathrm{~J}$
- $1200 \mathrm{~J}$
### Related Formula
Work done ($W$) in a thermodynamic process corresponds to the area under the curve on a $P-V$ diagram:
$$W = \int P \, dV$$
### Core Logic
Let's compute the work done along the distinct steps $AB$ and $BC$ matching the values shown in the graph:
From graph coordinates:
$$P_A = 8000 \mathrm{~Dyne/cm^2}, \quad V_A = 3 \mathrm{~m^3}$$
$$P_B = 4000 \mathrm{~Dyne/cm^2}, \quad V_B = 7 \mathrm{~m^3}$$
$$P_C = 4000 \mathrm{~Dyne/cm^2}, \quad V_C = 3 \mathrm{~m^3}$$
{{SOL_IMG1}}
### Step 1: Calculate Work Done in Process AB
Process $AB$ is a linear expansion. Fulfilling the area of the trapezoid underneath path $AB$:
$$W_{AB} = \frac{1}{2} (P_A + P_B) \Delta V_{AB}$$
$$W_{AB} = \frac{1}{2} (8000 + 4000) \mathrm{~Dyne/cm^2} \times (7 - 3) \mathrm{~m^3}$$
$$W_{AB} = 6000 \times 4 = 24000 \mathrm{~Dyne/cm^2} \cdot \mathrm{m^3}$$
### Step 2: Calculate Work Done in Process BC
Process $BC$ is an isobaric compression at $P = 4000 \mathrm{~Dyne/cm^2}$:
$$W_{BC} = P_B \cdot (V_C - V_B) = 4000 \times (3 - 7) = -16000 \mathrm{~Dyne/cm^2} \cdot \mathrm{m^3}$$
### Step 3: Total Work and Unit Conversion
Net work done is:
$$W_{\text{total}} = W_{AB} + W_{BC} = 24000 - 16000 = 8000 \mathrm{~Dyne/cm^2} \cdot \mathrm{m^3}$$
Converting units to Joules ($1 \mathrm{~N/m^2} = 10 \mathrm{~Dyne/cm^2} \implies 1 \mathrm{~Dyne/cm^2} = 0.1 \mathrm{~N/m^2}$):
$$W_{\text{total}} = 8000 \times (0.1 \mathrm{~N/m^2}) \cdot \mathrm{m^3} = 800 \mathrm{~J}$$
Since $800 \mathrm{~J}$ is not present in the options, this question was declared a **BONUS**.
### Pattern Recognition
Always be highly vigilant of mixed unit systems in thermodynamic graphics. Here, pressure is in CGS ($\text{Dyne/cm}^2$) while volume is in MKS ($\text{m}^3$). Converting early keeps you clear of algebraic traps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q1716 (2024)
Choose the correct statement for processes $\mathrm{A}$ & $\mathrm{B}$ shown in figure.
{{IMG1}}
- $\mathrm{PV}^{\gamma} = \mathrm{k} \text{ for process } \mathrm{B} \text{ and } \mathrm{PV} = \mathrm{k} \text{ for process } \mathrm{A}$
- $\mathrm{PV} = \mathrm{k} \text{ for process } \mathrm{B} \text{ and } \mathrm{A}$
- $\frac{\mathrm{P}^{\gamma - 1}}{\mathrm{T}^{\gamma}} = \mathrm{k} \text{ for process } \mathrm{B} \text{ and } \mathrm{T} = \mathrm{k} \text{ for process } \mathrm{A}$
- $\frac{\mathrm{T}^{\gamma}}{\mathrm{P}^{\gamma - 1}} = \mathrm{k} \text{ for process } \mathrm{A} \text{ and } \mathrm{PV} = \mathrm{k} \text{ for process } \mathrm{B}$
### Related Formula
$$\text{Isothermal: } PV = \text{constant}$$
$$\text{Adiabatic: } PV^{\gamma} = \text{constant}$$
$$\text{Adiabatic T-P relation: } T^{\gamma} P^{1-\gamma} = \text{constant} \implies \frac{T^{\gamma}}{P^{\gamma - 1}} = \text{constant}$$
### Core Logic
In a $P-V$ diagram starting from the same initial state and undergoing expansion, the adiabatic curve is steeper than the isothermal curve.
Looking at the graph, Curve (B) is steeper, so it represents an adiabatic process. Curve (A) is less steep, so it represents an isothermal process.
### Step 1: Check Options
Process A (Isothermal): $PV = k$ or $T = k$.
Process B (Adiabatic): $PV^{\gamma} = k$.
Option (1) states: $PV^{\gamma} = k$ for B and $PV = k$ for A. This is entirely correct.
Note: For an adiabatic process, $T^{\gamma} P^{1-\gamma} = k$, which can be written as $\frac{T^{\gamma}}{P^{\gamma-1}} = k$. Option (3) proposes $\frac{P^{\gamma-1}}{T^{\gamma}} = k$ for B, which is mathematically equivalent, but Option (1) is the standard and most direct NTA choice.
### Pattern Recognition
Slope of isothermal $= -P/V$. Slope of adiabatic $= -\gamma P/V$. Since $\gamma > 1$, adiabatic is always steeper. In expansions, adiabatic lies below isothermal.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q45 (2024)
Two thermodynamical process are shown in the figure. The molar heat capacity for process A and B are $C_A$ and $C_B$. The molar heat capacity at constant pressure and constant volume are represented by $C_P$ and $C_V$, respectively. Choose the correct statement.
{{IMG1}}
- $C_B = \infty , C_A = 0$
- $C_A = 0 \text{ and } C_B = \infty$
- $C_P > C_V > C_A = C_B$
- $C_A > C_P > C_V$
### Related Formula
$$C = C_v + \frac{R}{1 - n} \quad (\text{for } PV^n = \text{const})$$
$$\text{If } P = k V^x, \text{ then } n = -x$$
### Core Logic
For a generic straight line on a $\log P$ vs $\log V$ plot, the slope defines the exponent.
$\log P = x \log V + c \Rightarrow P = V^x \cdot e^c \Rightarrow PV^{-x} = \text{Constant}$.
The molar heat capacity is computed using $n = -x$.
### Step 1: Evaluate Process A
For process A: slope $= \gamma$ (given $\tan \theta > 1$).
$$\log P = \gamma \log V \Rightarrow PV^{-\gamma} = \text{Constant} \quad (\text{here } n = -\gamma)$$
$$C_A = C_V + \frac{R}{1 - (-\gamma)} = C_V + \frac{R}{1 + \gamma} \quad \dots (i)$$
### Step 2: Evaluate Process B
For process B: slope $= 1$ (given $\tan 45^\circ = 1$).
$$\log P = 1 \cdot \log V \Rightarrow PV^{-1} = \text{Constant} \quad (\text{here } n = -1)$$
$$C_B = C_v + \frac{R}{1 - (-1)} = C_v + \frac{R}{2} \quad \dots (ii)$$
### Step 3: Analyze Heat Capacities
We also know:
$$C_P = C_v + R \quad \dots (iii)$$
Comparing (i), (ii), and (iii):
Since $\gamma > 1$ for process A, $\frac{R}{1+\gamma} < \frac{R}{2}$.
Thus, $C_P > C_B > C_A > C_v$.
Note: Ans. By NTA (1 or 2). Ans. By our answer (Bonus). Since no option fully matches the mathematical derivation $C_P > C_B > C_A > C_v$, this was treated as a bonus or flawed option set. We assign index 0 based on official NTA preliminary keys marking 1 or 2.
### Pattern Recognition
Slopes on a log-log PV plot represent the polytropic exponent modifier. Positive slopes imply $-n > 0$, pulling heat capacity structurally between $C_v$ and $C_P$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q1988 (2024)
The given figure represents two isobaric processes for the same mass of an ideal gas, then
{{IMG1}}
- $\mathrm{P}_2\geq \mathrm{P}_1$
- $\mathrm{P}_2 > \mathrm{P}_1$
- $\mathrm{P_1 = P_2}$
- $\mathrm{P}_1 > \mathrm{P}_2$
### Related Formula
$PV = nRT$
### Core Logic
From the Ideal Gas Law:
$$V = \left(\frac{nR}{P}\right) T$$
In a $V-T$ graph, the equation of the line represents $y = mx$, where the slope $m$ is:
$$\text{Slope} = \frac{nR}{P}$$
$$\text{Slope} \propto \frac{1}{P}$$
Thus, a higher slope corresponds to a lower pressure.
### Step 2: Compare Slopes
From the given figure, the slope of line 2 is greater than the slope of line 1:
$$(\text{Slope})_2 > (\text{Slope})_1$$
Therefore, inversely:
$P_2 < P_1$
or $P_1 > P_2$.
### Pattern Recognition
In V-T graphs, steeper lines mean lower Pressure. In P-T graphs, steeper lines mean lower Volume. It's an inverse inverse slope relationship.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
