Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
A vessel contains 0.15 m$^3$ of a gas at pressure 8 bar and temperature 140 $^\circ$C with $c_p = 3R$ and $c_v = 2R$. It is expanded adiabatically till pressure falls to 1 bar. The work done during this process is __________ k J. (R is gas constant)
For adiabatic process, find final temperature: $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = \left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1}{2}$, so $T_2 = 206.5$ K
Find final volume using $P_1V_1^\gamma = P_2V_2^\gamma$: $V_2 = 0.6$ m³
Apply work formula for adiabatic process: $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$
5 moles of unknown gas is heated at constant volume from 10 °C to 20 °C. The molar specific heat of this gas at constant pressure $c_p = 8$ cal/mol.°C and $R = 8.36$ J/mol.°C. The change in the internal energy of the gas is _______ calorie.
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Answer: 300
Key idea: First Law of Thermodynamics and Mayer's Relation
Convert $R$ from J/mol·°C to cal/mol·°C: $R = \frac{8.36}{4.18} = 2$ cal/mol·°C
An ideal gas undergoes a process maintaining relation between pressure ($P$) and volume($V$) as $P = P_o \left[1 + \left(\frac{V_o}{V}\right)^2\right]^{-1}$, where $P_o$ and $V_o$ are constants. If two samples $A$ and $B$ (two moles each) with initial volumes $V_o$ and $3V_o$ respectively undergo above mentioned process and attain same pressure, then the difference at the temperatures of these samples, $T_B - T_A$ is _______.
($R$ = gas constant)
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Answer: 11P₀V₀/(10R)
Key idea: Ideal Gas Law with Non-standard Process
Write the pressure-volume relation: $P = P_o \left[1 + \left(\frac{V_o}{V}\right)^2\right]^{-1}$
For sample A at volume $V_o$: $P_A = \frac{P_o}{2}$. For sample B at volume $3V_o$: $P_B = \frac{9P_o}{10}$
Apply ideal gas law $PV = nRT$ with $n = 2$ moles for both samples
Given below are two statements: one is labelled as **Assertion A** and the other is labelled as **Reason R**
**Statement I:** Change in internal energy of a system containing $n$ mole of ideal gas can be written as $\Delta U = n C_V (T_f - T_i) = \frac{nR}{\gamma - 1}(T_f - T_i)$, where $\gamma = \frac{C_p}{C_V}$, $T_i$ = initial temperature, $T_f$ = final temperature.
**Statement II:** Relation between degree of freedom $f$ and $\gamma (= C_p/C_v)$ is $\gamma = 1 + \frac{2}{f}$
Choose the *correct* answer from the options given below
(1) Both A and R are true and R is the correct explanation of A
(2) Both A and R are true but R is NOT the correct explanation of A
(3) A is true but R is false
(4) A is false but R is true
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Relationship between heat capacities, gamma, and degrees of freedom for ideal gases
Consider the following statements:
A. Zeroth law of thermodynamics gives concept of temperature
B. First law of thermodynamics gives concept of internal energy
C. In isothermal expansion of ideal gas, $\Delta Q \neq \Delta W$
D. Product of intensive and extensive variables is extensive
E. The ratio of any extensive variable to mass will be an extensive variable
Choose the correct combination of statements from the options given below:
(1) C, D and E Only
(2) A, B and C Only
(3) A, B and D Only
(4) B, C and D Only
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Answer: Option (3)
Key idea: Laws of Thermodynamics and Thermodynamic Variables
**Statement A**: Zeroth law establishes thermal equilibrium and gives the concept of temperature. ✓ TRUE
**Statement B**: First law ($\Delta Q = \Delta U + \Delta W$) introduces internal energy as a state function. ✓ TRUE
**Statement C**: For isothermal process, $\Delta T = 0 \Rightarrow \Delta U = 0$, so $\Delta Q = \Delta W$ (not $\neq$). ✗ FALSE
**Statement D**: Intensive × Extensive = Extensive (e.g., $P \times V = PV$ is extensive). ✓ TRUE
**Statement E**: Extensive/Extensive = Intensive (e.g., Volume/mass = specific volume is intensive). ✗ FALSE
A certain gas is isothermally compressed to $\left(\frac{1}{3}\right)^{rd}$ of its initial volume ($V_{o} = 3$ litre) by applying required pressure. If the bulk modulus of the gas is $3 \times 10^{5}$ N/m$^{2}$, the magnitude of work done on the gas is ____J.
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Answer: 989
Key idea: Isothermal compression work and bulk modulus relationship
For isothermal process, bulk modulus $B = P$, so initial pressure $P_0 = 3 \times 10^5$ N/m²
A cylinder with adiabatic walls is closed at both ends and is divided into two compartments by a frictionless adiabatic piston. Ideal gas is filled in both (left and right) the compartments at same $P$, $V$, $T$. Heating is started from left side until pressure changes to $\frac{27P}{8}$. If initial volume of each compartment was 9 litres then the final volume in right-hand side compartment is _______ litres. (for this ideal gas $C_P/C_V = 1.5$)
(1) 3
(2) 4
(3) 14
(4) 9
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Answer: Option (2)
Key idea: Adiabatic process for ideal gas with moving piston
Identify that the right compartment undergoes adiabatic compression (adiabatic walls, no heating). The piston ensures equal pressure on both sides.
Write the adiabatic equation for the right side: $PV^\gamma = P_fV_f^\gamma$ where $\gamma = 3/2$
One mole of diatomic gas having rotational modes only is kept in a cylinder with a piston system. The cross-section area of the cylinder is 4 cm$^2$. The gas is heated slowly to raise the temperature by 1.2 $^\circ$C during which the piston moves by 25 mm. The amount of heat supplied to the gas is ______ J. (Atmospheric pressure = 100 kPa, $R = 8.3$ J/mol. K) (Neglect mass of the piston)
(1) 24.8
(2) 25
(3) 15.04
(4) 29.98
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Answer: Option (2)
Key idea: First Law of Thermodynamics for Isobaric Process
Identify this as an isobaric process (constant pressure) since the piston moves freely under atmospheric pressure
For diatomic gas with rotational modes: $f = 5$ (3 translational + 2 rotational), so $C_v = \frac{5}{2}R = 2.5R$
Calculate change in internal energy: $\Delta U = nC_v\Delta T = 1 \times 2.5 \times 8.3 \times 1.2 = 24.9$ J
Calculate work done: $W = P\Delta V = P \times A \times d = 10^5 \times 4 \times 10^{-4} \times 0.025 = 1$ J
Apply First Law: $Q = \Delta U + W = 24.9 + 1 = 25.9 \approx 25$ J
Initial pressure and volume of a monoatomic ideal gas are $P$ and $V$. The change in internal energy of this gas in adiabatic expansion to volume $V_{final} = 27$ $V$ is ______ J.
(1) $-2PV(3\sqrt{3}-1)$
(2) $\frac{4}{3}PV$
(3) $-\frac{4}{3}PV$
(4)
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Answer: Option (3)
Key idea: Adiabatic process for ideal gas, change in internal energy using PV relationship
For monoatomic ideal gas: $\gamma = \frac{5}{3}$, so $\gamma - 1 = \frac{2}{3}$
