JEE Physics Thermodynamics MCQs Set 23

Answer: (a) \( 600R \)

Explanation:
1. Heats: AB \( =nC_p\Delta T=2(2.5R)(300)=1500R \); BC \( =nRT_B\ln2=1200R\ln2 \); CD \( =nC_V\Delta T=2(1.5R)(-300)=-900R \); DA \( =nRT_A\ln\dfrac14=-1200R\ln2 \).
2. For a cycle \( W=Q_{net}=1500R-900R=600R \) (the logarithmic terms cancel).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) \( 1200R \)

Explanation:
1. On AB \( p=\dfrac cT \) and \( V=\dfrac{nRT}{p}=\dfrac{nRT^2}{c} \), so \( dV=\dfrac{2nRT}{c}dT \).
2. Work by the gas \( =\int p\,dV=2nR\,\Delta T=2(2)R(T_1-2T_1)=-1200R \), so the work done on the gas is \( 1200R \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) \( 1500R \)

Explanation:
1. Isobaric heating: \( Q=nC_p\Delta T \).
2. \( Q=2\times\tfrac52R\times300=1500R \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) \( \dfrac32p_1V_1\left[1-\left(\dfrac{V_1}{V_2}\right)^{2/3}\right] \)

Explanation:
1. Adiabatic compression: \( p_2=p_1\left(\dfrac{V_1}{V_2}\right)^{5/3} \), and the work done by the gas \( =\dfrac{p_1V_1-p_2V_2}{\gamma-1}=\dfrac32p_1V_1\left[1-\left(\dfrac{V_1}{V_2}\right)^{2/3}\right] \), which is negative since \( V_2\lt V_1 \).
2. No work is done in the constant volume heating, so this is the total work.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) \( \dfrac{p_1V_1^{5/3}V_2^{-2/3}}{2R}+\dfrac{Q}{3R} \)

Explanation:
1. After the adiabatic compression \( p_2V_2=p_1V_1^{5/3}V_2^{-2/3} \), so \( T_2=\dfrac{p_2V_2}{nR}=\dfrac{p_1V_1^{5/3}V_2^{-2/3}}{2R} \).
2. At constant volume \( Q=nC_V\Delta T=2(1.5R)\Delta T\Rightarrow\Delta T=\dfrac{Q}{3R} \), so \( T_f=T_2+\dfrac{Q}{3R} \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) \( \dfrac52p_0V_0\text{ and }3p_0V_0 \)

Explanation:
1. CA is isobaric compression: \( Q=nC_p\Delta T=\tfrac52p_0(V_0-2V_0)=-\tfrac52p_0V_0 \), i.e. \( \tfrac52p_0V_0 \) is rejected.
2. AB is isochoric: \( Q=\tfrac32(3p_0V_0-p_0V_0)=3p_0V_0 \), absorbed.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) \( \dfrac12p_0V_0 \)

Explanation:
1. Net heat in the cycle \( =W=p_0V_0=Q_{AB}+Q_{BC}+Q_{CA}=3p_0V_0+Q_{BC}-\tfrac52p_0V_0 \).
2. \( Q_{BC}=p_0V_0-\tfrac12p_0V_0=\tfrac12p_0V_0 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) 481 K

Explanation:
1. \( n=\dfrac{2000}{4}=500 \) mol.
2. \( T_C=\dfrac{p_CV_C}{nR}=\dfrac{10\times10^4\times20}{500\times8.31}\approx481 \) K (and \( T_A\approx120 \) K, \( T_B\approx241 \) K).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) No, because both samples end in the same state and the internal energy is a state function

Explanation:
1. Heat and work are path functions, but the final state (pressure, volume, temperature, internal energy) of the gas depends only on the end point C.
2. Nothing in the final state records the path, so the two samples cannot be told apart.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) \( 3.25\times10^6\ \text{J} \)

Explanation:
1. AB (isochoric): \( Q=\tfrac32(p_BV-p_AV)=1.5(10^6-5\times10^5)=7.5\times10^5 \) J.
2. BC (isobaric): \( Q=\tfrac52p\Delta V=2.5\times10^5\times10=2.5\times10^6 \) J. Total \( =3.25\times10^6 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) 909 K and 791 K

Explanation:
1. \( T_B=T_A\left(\dfrac{V_A}{V_B}\right)^{0.4}=300\times16^{0.4}\approx909 \) K. Isobaric BC: \( T_C=2T_B=1818 \) K.
2. \( V_D=V_A=8V_C \) (since \( V_C=2V_B \) and \( V_A=16V_B \)): \( T_D=T_C\left(\dfrac{V_C}{V_D}\right)^{0.4}=1818\times8^{-0.4}\approx791 \) K.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 61%

Explanation:
1. Heat absorbed in BC: \( Q_1=C_p(T_C-T_B)=\tfrac72R(909)=3181R \). Heat rejected in DA: \( Q_2=C_V(T_D-T_A)=\tfrac52R(491)=1228R \).
2. \( \eta=1-\dfrac{Q_2}{Q_1}=1-\dfrac{1228}{3181}\approx0.61 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) 2.77 kJ

Explanation:
1. \( T_2=300\left(\dfrac12\right)^{2/3}\approx189 \) K.
2. \( W=-\Delta U=nC_V(T_1-T_2)=2(1.5)(8.31)(111)\approx2.77 \) kJ.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) 10.8%

Explanation:
1. Heat absorbed \( =5960+3645=9605 \) J; net work \( =1040 \) J.
2. \( \eta=\dfrac{1040}{9605}\approx10.8\% \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 1.87 kJ

Explanation:
1. Adiabatic: \( p^{1-\gamma}T^{\gamma}= \) const gives \( \dfrac{T_B}{T_A}=\left(\dfrac{p_B}{p_A}\right)^{(\gamma-1)/\gamma}=\left(\dfrac23\right)^{2/5}=0.85 \), so \( T_B=850 \) K.
2. \( W_{AB}=nC_V(T_A-T_B)=1.5\times8.31\times150\approx1870 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) 5.3 kJ

Explanation:
1. BC is at constant volume: \( \dfrac{T_C}{T_B}=\dfrac{p_C}{p_B}=\dfrac12\Rightarrow T_C=425 \) K.
2. \( Q=nC_V(T_B-T_C)=1.5\times8.31\times425\approx5.3 \) kJ lost.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) 1152 J

Explanation:
1. Work: AB \( =nR(100)=200R \); BC (isothermal, volume doubles) \( =nR(400)\ln2=800R\ln2 \); CD \( =-200R \); DA \( =nR(300)\ln\dfrac12=-600R\ln2 \).
2. Net \( W=200R\ln2=200\times8.31\times0.693\approx1152 \) J (equal to the net heat, and \( \Delta U=0 \)).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) \( 3RT_A\left(\ln2-\dfrac12\right) \)

Explanation:
1. Isothermal expansion: \( W_1=3RT_A\ln2 \). Isobaric compression at pressure \( \dfrac{p_A}{2} \) from 2V to V: \( W_2=-\dfrac{p_A}{2}V_A=-\dfrac{3RT_A}2 \). Isochoric step: \( W_3=0 \).
2. Net work \( =3RT_A\ln2-\dfrac32RT_A \). Since \( \Delta U=0 \) over a cycle, the net heat supplied equals this.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (a) 5

Explanation:
1. \( TV^{\gamma-1}= \) const: \( 5.66^{\gamma-1}=2 \). Since \( 5.66=4\sqrt2=2^{5/2} \), \( \tfrac52(\gamma-1)=1\Rightarrow\gamma=1.4 \).
2. \( \gamma=1+\dfrac2f=1.4\Rightarrow f=5 \).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (c) 675 K and \( 3.6\times10^6 \) N/m\(^2\)

Explanation:
1. \( n=\dfrac{pV}{RT}=\dfrac{1.6\times10^6\times0.0083}{8.3\times300}=5.33 \) mol. \( C_p=\tfrac52R \) means a monoatomic gas, so \( C_V=\tfrac32R \).
2. Closed vessel: \( Q=nC_V\Delta T\Rightarrow\Delta T=\dfrac{2.49\times10^4}{5.33\times12.45}=375 \) K, so \( T=675 \) K and \( p'=p\dfrac{675}{300}=3.6\times10^6 \) N/m\(^2\).

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (d) 972 J

Explanation:
1. \( C_V=\tfrac32R\Rightarrow\gamma=\tfrac53 \). \( p_2=10^5\left(\dfrac62\right)^{5/3}=6.24\times10^5 \) N/m\(^2\).
2. Work done on the gas \( =\dfrac{p_2V_2-p_1V_1}{\gamma-1}=\dfrac{1248-600}{2/3}\approx972 \) J.

Teacher's Note:
1. First write the first law, Q = delta U + W, and recall that delta U depends only on the temperature change (n Cv delta T) for an ideal gas, whatever the path.
2. Work is the area under the p-V curve (positive for expansion); for a cycle the net work equals the net heat absorbed.

Answer: (b) 1 N/m\(^2\)

Explanation:
1. Each elastic collision reverses the momentum: change \( =2mv=2\times10^{-26}\times10^4=2\times10^{-22} \) kg m/s.
2. Force on 1 m\(^2\) \( =10^{22}\times2\times10^{-22}=2 \) N, so the pressure is 2 N/m\(^2\), of the order of 1 N/m\(^2\).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (c) 25

Explanation:
1. Linear scale: \( \dfrac{x-x_0/3}{x_0-x_0/3}=\dfrac{T-0}{100-0} \).
2. \( \dfrac{x_0/2-x_0/3}{2x_0/3}=\dfrac14=\dfrac T{100}\Rightarrow T=25^\circ \)C.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) 90 W m\(^{-2}\)

Explanation:
1. Energy flux \( =K\dfrac{\Delta T}{L} \).
2. \( =0.1\times\dfrac{1000-100}{1}=90 \) W m\(^{-2}\).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) \( \dfrac{\gamma+1}{2} \)

Explanation:
1. Mean time \( \tau=\dfrac\lambda{v_{rms}} \) with \( \lambda\propto V \) (fixed number of molecules) and \( v_{rms}\propto\sqrt T \).
2. Adiabatic: \( T\propto V^{1-\gamma} \), so \( v_{rms}\propto V^{(1-\gamma)/2} \) and \( \tau\propto V^{1-(1-\gamma)/2}=V^{(\gamma+1)/2} \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (c) \( \ln2,\ \ln2 \)

Explanation:
1. Entropy is a state function: the change in the body depends only on the initial and final temperatures, not on the number of reservoirs used.
2. \( \Delta S=\int_{100}^{200}\dfrac{C\,dT}{T}=1\times\ln\dfrac{200}{100}=\ln2 \) in both cases.

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (d) \( \dfrac1{2\pi}\sqrt{\dfrac{A^2\gamma p_0}{MV_0}} \)

Explanation:
1. Isolated gas, so the change is adiabatic: \( pV^\gamma= \) const gives \( dp=-\gamma\dfrac{p_0}{V_0}dV \) with \( dV=A\,x \).
2. Restoring force \( F=-A\,dp=-\dfrac{\gamma p_0A^2}{V_0}x \), so \( \omega=\sqrt{\dfrac{\gamma p_0A^2}{MV_0}} \) and \( f=\dfrac{1}{2\pi}\sqrt{\dfrac{A^2\gamma p_0}{MV_0}} \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.

Answer: (b) \( \dfrac3T \)

Explanation:
1. \( p=\dfrac{c}{T^2} \) and \( V=\dfrac{nRT}{p}=\dfrac{nRT^3}{c}\propto T^3 \).
2. \( \gamma_V=\dfrac1V\dfrac{dV}{dT}=\dfrac3T \).

Teacher's Note:
1. Break the problem into the simplest physical law that applies (kinetic theory, linear scale of a thermometer, conduction or radiation) and write it before substituting numbers.
2. Check orders of magnitude and units at the end; most mistakes in these problems come from a missing factor of 2 or of 4 pi.