Thermodynamics Previous Year Questions — JEE Main Physics
36 past-year Thermodynamics questions from JEE Main (Physics).
Q10 (2025)
Identify the characteristics of an adiabatic process in a monoatomic gas.
(A) Internal energy is constant.
(B) Work done in the process is equal to the change in internal energy.
(C) The product of temperature and volume is a constant.
(D) The product of pressure and volume is a constant.
(E) The work done to change the temperature from $T_1$ to $T_2$ is proportional to $(T_2 - T_1)$
Choose the correct answer from the options given below:
- $\text{(A), (C), (D) only}$
- $\text{(A), (C), (E) only}$
- $\text{(B), (E) only}$
- $\text{(B), (D) only}$
### Related Formula
1. First Law of Thermodynamics:
$dQ = dU + dW$
In an adiabatic process:
$$dQ = 0 \implies dW = -dU$$
2. Change in Internal Energy:
$$dU = n C_v dT = n C_v (T_2 - T_1)$$
### Core Logic
Let's analyze each statement:
- **(A) Internal energy is constant:** Incorrect. Since temperature changes during an adiabatic expansion/compression, internal energy ($U \propto T$) must change.
- **(B) Work done is equal to the change in internal energy:** Correct in magnitude ($|dW| = |dU|$). By definition, $dW = -dU$, which correlates the magnitude of work to the change in internal energy.
- **(C) Product of temperature and volume is constant:** Incorrect. The adiabatic equation of state is $T V^{\gamma-1} = \text{constant}$.
- **(D) Product of pressure and volume is constant:** Incorrect. The relation is $P V^\gamma = \text{constant}$.
- **(E) Work done is proportional to $(T_2 - T_1)$:** Correct. Since $dW = -dU = -n C_v (T_2 - T_1)$, work done is directly proportional to the temperature change $(T_2 - T_1)$.
### Step 1: Determine the correct option
Since only statements (B) and (E) are correct, the correct option is (3).
### Pattern Recognition
Sees: Characteristics of adiabatic thermodynamic process.
Trap: Confusing adiabatic state relations ($PV^\gamma = C$, $TV^{\gamma-1} = C$) with isothermal state relations ($PV = C$, $T = C$).
Shortcut: First law of thermodynamics under $dQ=0$ strictly enforces $|dW| = |dU|$, which validates statement B and E immediately.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q12 (2025)
In an adiabatic process, which of the following statements is true?
- The molar heat capacity is infinite
- Work done by the gas equals the increase in internal energy
- The molar heat capacity is zero
- The internal energy of the gas decreases as the temperature increases
### Related Formula
$dQ = n C dT$
$$dQ = 0 \quad \text{(for adiabatic process)}$$
### Core Logic
An adiabatic process involves no heat exchange between the system and its surroundings ($dQ = 0$).
The molar heat capacity $C$ is defined as:
$$C = \frac{1}{n}\frac{dQ}{dT}$$
Since $dQ = 0$ while the temperature changes ($dT \neq 0$):
$C = 0$
Thus, the molar heat capacity for any adiabatic process is always zero.
### Step 1: Check other options
- Option (1): Isothermal processes have infinite molar heat capacity ($dT = 0$).
- Option (2): From the First Law ($dQ = dU + dW \implies dW = -dU$), the work done equals the *decrease* in internal energy.
- Option (4): The internal energy of an ideal gas ($dU = n C_{\mathrm{v}} dT$) increases directly as temperature increases.
### Step 2: Final Conclusion
The statement 'The molar heat capacity is zero' is true.
### Pattern Recognition
Adiabatic = no heat flow ($dQ=0$). Since molar heat capacity tracks the ratio of heat input to temperature change, $C$ must be 0. Conversely, isothermal has infinite capacity because heat is absorbed without any temperature change ($dT=0$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q155 (2025)
An ideal gas exists in a state with pressure $P_{0}$, volume $V_{0}$. It is isothermally expanded to 4 times of its initial volume $(V_{0})$, then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is :
- $P_{0}V_{0}(2\ln 2-0.75)$
- $P_{0}V_{0}(\ln 2-0.75)$
- $P_{0}V_{0}(\ln 2-0.25)$
- $P_{0}V_{0}(2\ln 2-0.25)$
### Related Formula
For a cyclic thermodynamic process, the net change in internal energy is zero:
$$\Delta U_{\text{cyclic}} = 0$$
By the First Law of Thermodynamics, the total heat exchanged $Q_T$ equals the net work done $W_{\text{net}}$:
$$Q_T = W_{\text{net}} = W_1 + W_2 + W_3$$
### Core Logic
The cycle consists of three steps:
1. Isothermal expansion from $(P_0, V_0)$ to volume $4V_0$.
2. Isobaric compression to the original volume $V_0$.
3. Isochoric heating back to the initial state.
{{SOL_IMG1}}
### Step 1: Work in Isothermal Expansion ($W_1$)
Initial state: $(P_0, V_0)$. Final state volume: $4V_0$.
$$W_1 = P_0 V_0 \ln\left(\frac{4V_0}{V_0}\right) = P_0 V_0 \ln(4) = 2 P_0 V_0 \ln(2)$$
Also, the pressure at the end of this process is:
$$P_1 = \frac{P_0 V_0}{4V_0} = \frac{P_0}{4}$$
### Step 2: Work in Isobaric Compression ($W_2$)
The process occurs at constant pressure $P = P_1 = P_0/4$. The volume goes from $4V_0$ back to $V_0$:
$$W_2 = P \Delta V = \frac{P_0}{4} (V_0 - 4V_0) = \frac{P_0}{4} (-3V_0) = -0.75 P_0 V_0$$
### Step 3: Work in Isochoric Heating ($W_3$)
Since the volume is held constant at $V_0$, no boundary work is done:
$W_3 = 0$
### Step 4: Total Heat Exchanged ($Q_T$)
$$Q_T = W_{\text{net}} = W_1 + W_2 + W_3$$
$$Q_T = 2 P_0 V_0 \ln(2) - 0.75 P_0 V_0 = P_0 V_0 (2\ln(2) - 0.75)$$
### Pattern Recognition
In any cyclic system returning to its initial state, finding total heat is mathematically equivalent to calculating the enclosed area on a P-V diagram. Here, the isobaric step occurs at the lowest expanded pressure, resulting in a simple negative rectangular area correction subtracted from the logarithmic isothermal expansion curve.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q250 (2025)
An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is $\times 10^{-1}$ J. (Take $\pi = 3.14$ ) {{IMG1}}
### Related Formula
The work done $W$ in a cyclic thermodynamic process is equal to the area enclosed by the loop on a Pressure-Volume ($P-V$) diagram:
$$W = \text{Area of Closed Loop}$$
For an ellipse with semi-major axis $a$ and semi-minor axis $b$:
$$\text{Area} = \pi a b$$
### Core Logic
Determine the semi-axes of the elliptical cycle on the $P-V$ plane:
- On the Pressure axis (x-axis):
$$a = \frac{P_{\text{max}} - P_{\text{min}}}{2} = \frac{500 - 300}{2} \mathrm{~kPa} = 100 \mathrm{~kPa} = 10^5 \mathrm{~Pa}$$
- On the Volume axis (y-axis):
$$b = \frac{V_{\text{max}} - V_{\text{min}}}{2} = \frac{350 - 150}{2} \mathrm{~cm}^3 = 100 \mathrm{~cm}^3 = 100 \times 10^{-6} \mathrm{~m}^3 = 10^{-4} \mathrm{~m}^3$$
### Step 1: Calculate Area
Substitute $a$ and $b$ in standard SI units into the area equation:
$$W = \pi a b = 3.14 \times (10^5 \mathrm{~Pa}) \times (10^{-4} \mathrm{~m}^3)$$
$$W = 3.14 \times 10 = 31.4 \mathrm{~J}$$
Express in terms of $\times 10^{-1} \mathrm{~J}$:
$$W = 314 \times 10^{-1} \mathrm{~J}$$
Thus, the multiplier is $314$.
### Pattern Recognition
Sees: Circular/elliptical thermodynamic cycle.
Shortcut: Work done is always $\pi \Delta P \Delta V / 4$. Simply compute the semi-axes difference $100 \mathrm{~kPa}$ and $100 \mathrm{~cm^3}$ and multiply by $\pi$ directly, keeping tracking of metric prefixes ($10^3 \times 10^{-6} = 10^{-3}$).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Q5 (2025)
Water falls from a height of $200\mathrm{~m}$ into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take $g = 10\mathrm{~m/s}^{2}$, specific heat of water $= 4200\mathrm{~J/(kg\cdot K)}$)
- $0.23\mathrm{~K}$
- $0.36\mathrm{~K}$
- $0.14\mathrm{~K}$
- $0.48\mathrm{~K}$
### Related Formula
$$\Delta U = mgh \quad \text{and} \quad Q = ms\Delta T$$
By conservation of energy (assuming all potential energy goes into heating the water):
$$mgh = ms\Delta T \implies \Delta T = \frac{gh}{s}$$
where,
$g$ = acceleration due to gravity
$h$ = height of the fall
$s$ = specific heat of water
$\Delta T$ = rise in temperature
### Core Logic
Given parameters:
- $h = 200\mathrm{~m}$
- $g = 10\mathrm{~m/s}^{2}$
- $s = 4200\mathrm{~J/(kg\cdot K)}$
Substitute the values to find $\Delta T$:
$$\Delta T = \frac{10 \times 200}{4200} = \frac{2000}{4200}$$
$$\Delta T = \frac{10}{21} \approx 0.476\mathrm{~K} \approx 0.48\mathrm{~K}$$
### Pattern Recognition
Sees: "Water falling from height $h$ raises temperature" → Mass cancels out. $\Delta T = \frac{gh}{s}$.
Shortcut: Always use SI units ($s_{\text{water}} = 4200\mathrm{~J/kg\cdot K}$ is given; if given in $\mathrm{cal/g\cdot^\circ C}$, convert using $1\mathrm{~cal} = 4.184\mathrm{~J}$). ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Thermodynamics
Class 11 Physics: Work, Energy and Power
