Thermodynamics — Top 20 Questions

20Questions

0Correct

0Attempted

During an adiabatic process pressure \(\propto T^3\). Ratio \(C_p/C_v\):

A5/3

B3/2

C7/5

D9/7

✅ Correct: B

Adiabatic: \(PT^{-\gamma/(\gamma-1)} = \text{const}\). Given \(P \propto T^3\) → exponent \(= -\gamma/(\gamma-1) \Rightarrow -3\). Solving: \(\gamma/(\gamma-1) = 3\) → \(\gamma = 3/2\). So \(C_p/C_v = \mathbf{3/2}\).

Pt resistance thermometer: \(R_0 = 8\Omega\), \(R_{100} = 10\Omega\). Resistance at 400°C:

A10 Ω

B16 Ω

C8 Ω

D12 Ω

✅ Correct: B

\(R_t = R_0(1 + \alpha t)\). \(\alpha = (R_{100}-R_0)/(100 R_0) = 2/800 = 1/400\). \(R_{400} = 8(1 + 400/400) = 8\times2 = \mathbf{16\Omega}\).

Metal sheet with hole (diameter 5 cm at 27°C). Heated to 177°C with \(\alpha = 1.6\times10^{-5}\) °C⁻¹. Change in diameter (cm):

A1.2×10⁻³

B1.6×10⁻³

C2.4×10⁻³

D0.8×10⁻³

✅ Correct: A

\(\Delta d = d_0 \alpha \Delta T = 5 \times 1.6\times10^{-5} \times 150 = 5\times2.4\times10^{-3} = 1.2\times10^{-2}\)... \(= 5\times1.6\times10^{-5}\times150 = \mathbf{1.2\times10^{-2}}\) cm. Hole expands like solid.

Speed of sound in oxygen at STP (\(R = 8.3\) J mol⁻¹K⁻¹, \(\gamma = 1.4\), \(M_{O_2} = 32\times10^{-3}\) kg):

A341 m/s

B333 m/s

C325 m/s

D315 m/s

✅ Correct: D

\(v = \sqrt{\gamma RT/M} = \sqrt{1.4\times8.3\times273/0.032} = \sqrt{99218} \approx \mathbf{315\text{ m/s}}\).

At what temperature is \(v_{rms}\) of H₂ equal to that of O₂ at 47°C?

A20 K

B80 K

C4 K

D−73 K

✅ Correct: A

\(v_{rms} = \sqrt{3RT/M}\). Equal → \(T_{H_2}/M_{H_2} = T_{O_2}/M_{O_2}\) → \(T_{H_2} = 320\times2/32 = \mathbf{20\text{ K}}\).

Temperature at which average KE of O₂ is double its value at 27°C:

A627°C

B927°C

C327°C

D54°C

✅ Correct: C

\(\langle KE \rangle = \frac{3}{2}kT\). Double → \(T' = 2T = 2\times300 = 600\text{ K} = \mathbf{327°C}\).

In a Carnot engine \(T_H = 600\) K and \(T_C = 300\) K. Efficiency:

A25%

B33%

C50%

D66%

✅ Correct: C

\(\eta = 1 - T_C/T_H = 1 - 300/600 = 0.5 = \mathbf{50\%}\).

First law of thermodynamics \(Q = \Delta U + W\). For adiabatic process:

A\(Q = W\)

B\(\Delta U = -W\)

C\(Q = \Delta U\)

D\(W = 0\)

✅ Correct: B

Adiabatic: \(Q = 0\). So \(0 = \Delta U + W \Rightarrow \mathbf{\Delta U = -W}\). Internal energy decreases when work is done by the gas.

An ideal gas expands isothermally from \(V\) to \(2V\). Work done by gas:

A\(nRT\ln 2\)

B\(nRT\)

C\(nRT/2\)

✅ Correct: A

\(W = nRT\ln(V_f/V_i) = nRT\ln(2V/V) = \mathbf{nRT\ln 2}\). For isothermal process, internal energy change = 0, so \(Q = W\).

Q10

For an ideal gas, in an isobaric process from \(T_1\) to \(T_2\), work done:

A\(nC_v(T_2-T_1)\)

B\(nR(T_2-T_1)\)

C\(nC_p(T_2-T_1)\)

✅ Correct: B

Isobaric: \(W = P\Delta V = nR\Delta T = \mathbf{nR(T_2-T_1)}\). Also \(Q = nC_p\Delta T\) and \(\Delta U = nC_v\Delta T\).

Q11

A gas with \(\gamma = 5/3\) is compressed adiabatically to half its volume. Pressure ratio \(P_f/P_i\):

A\(2^{5/3}\)

C\(2^{3/5}\)

✅ Correct: A

Adiabatic: \(PV^\gamma = \text{const}\). \(P_i V^\gamma = P_f (V/2)^\gamma\). \(P_f/P_i = 2^\gamma = 2^{5/3} = \mathbf{2^{5/3}}\).

Q12

Thermal efficiency of Carnot engine = 40%. Heat source at 500 K. Sink temperature:

A200 K

B300 K

C400 K

D250 K

✅ Correct: B

\(\eta = 1 - T_C/T_H\) → \(0.4 = 1 - T_C/500\) → \(T_C = 500\times0.6 = \mathbf{300\text{ K}}\).

Q13

Entropy change in isothermal expansion of ideal gas from \(V\) to \(2V\):

A\(nR\ln 2\)

C\(-nR\ln 2\)

D\(2nR\)

✅ Correct: A

\(\Delta S = Q/T = nRT\ln 2/T = \mathbf{nR\ln 2}\). Entropy increases in irreversible isothermal expansion.

Q14

Zeroth law of thermodynamics defines:

ATemperature

BEntropy

CInternal energy

DHeat capacity

✅ Correct: A

Zeroth law: if A and C are in thermal equilibrium, and B and C are in thermal equilibrium, then A and B are in equilibrium. This defines the concept of temperature as an intrinsic property.

Q15

A real gas obeys \((P + a/V^2)(V-b) = RT\). At very high temperature and large volume, it behaves as:

ASolid

BIdeal gas

CLiquid

DPlasma

✅ Correct: B

At high \(T\) and large \(V\): \(a/V^2 \to 0\) and \(b/V \to 0\). Equation reduces to \(PV = RT\) → behaves as ideal gas.

Q16

Heat required to raise 1 mole monoatomic ideal gas from \(T\) to \(2T\) at constant volume (\(R = 8.3\) J/mol K):

A\(\frac{3}{2}RT\)

B\(\frac{5}{2}RT\)

C\(3RT\)

D\(2RT\)

✅ Correct: A

\(Q = nC_v\Delta T = 1 \times \frac{3}{2}R \times T = \mathbf{\frac{3}{2}RT}\). For monoatomic gas \(C_v = \frac{3}{2}R\).

Q17

In a PV diagram, area enclosed by a closed cycle represents:

AChange in internal energy

BNet heat absorbed

CNet work done

DChange in entropy

✅ Correct: C

Area enclosed in PV diagram = \(\oint P\,dV\) = net work done by gas in one cycle. For clockwise cycle, work is positive (done by gas). For anticlockwise, work is done on gas.

Q18

Temperature at which Celsius and Fahrenheit scales give the same reading:

A40°

B−40°

C−273°

D100°

✅ Correct: B

\(C = (F-32)\times5/9\). Setting \(C = F\): \(F = (F-32)\times5/9\) → \(9F = 5F - 160\) → \(4F = -160\) → \(F = C = \mathbf{-40°}\).

Q19

For which process is \(\Delta U = 0\)?

AAdiabatic

BIsothermal

CIsobaric

DIsochoric

✅ Correct: B

For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature is constant → \(\Delta U = nC_v\Delta T = 0\).

Q20

An engine takes 10000 J from heat source and gives 4000 J as useful work. Heat rejected to sink:

A6000 J

B4000 J

C10000 J

D14000 J

✅ Correct: A

First law for engine: \(Q_H = W + Q_C\). \(Q_C = 10000 - 4000 = \mathbf{6000\text{ J}}\). Efficiency \(= 4000/10000 = 40\%\).

Roshan

Expert · 5 Years Experience

Top 20 questions from PYQ analysis. NeetJeeRankers pe free expert-level practice milti hai.

Home | Physics Syllabus