JEE Physics Thermodynamics MCQs Set 17

Answer: (c) \( C_p-C_v=\dfrac{R}{28} \)

Explanation:
1. For one mole, \( C_P-C_V=R \); per unit mass divide by the molar mass.
2. \( M_{N_2}=28 \), so \( C_p-C_v=\dfrac R{28} \).

Teacher's Note:
1. Specific heat per unit mass is the molar value divided by the molar mass.
2. Do not confuse molar and specific heats.

Answer: (a) 90 J

Explanation:
1. As an engine \( \eta=1-\dfrac{T_2}{T_1}=\dfrac1{10} \), so \( \dfrac{T_2}{T_1}=0.9 \).
2. Coefficient of performance \( \beta=\dfrac{T_2}{T_1-T_2}=\dfrac{0.9}{0.1}=9 \).
3. Heat absorbed from the cold reservoir \( =\beta W=9\times10=90 \) J.

Teacher's Note:
1. Refrigerator coefficient of performance is \( \dfrac{1-\eta}{\eta} \) for the same Carnot cycle.
2. Heat extracted equals COP times work input.

Answer: (d) 6 cal

Explanation:
1. Along iaf: \( \Delta U=Q-W=50-20=30 \) cal.
2. Along ibf: \( W=Q-\Delta U=36-30=6 \) cal.

Teacher's Note:
1. \( \Delta U \) is the same on every path between i and f.
2. Apply the first law separately on each path.

Answer: (d) \( 5\times10^{4} \) J

Explanation:
1. Volume \( V=\dfrac{m}{\rho}=\dfrac14=0.25\ \text{m}^3 \).
2. Thermal (internal) energy of a diatomic gas \( =\tfrac52PV=\tfrac52\times8\times10^{4}\times0.25 \).
3. \( =5\times10^{4} \) J.

Teacher's Note:
1. Internal energy of an ideal gas is \( \tfrac f2PV \).
2. Diatomic: \( f=5 \) at ordinary temperatures.

Answer: (b) 372 K and 310 K

Explanation:
1. \( 1-\dfrac{T_2}{T_1}=\dfrac16 \Rightarrow T_2=\dfrac56T_1 \).
2. \( 1-\dfrac{T_2-62}{T_1}=\dfrac13 \Rightarrow T_2-62=\dfrac23T_1 \).
3. \( \dfrac56T_1-\dfrac23T_1=62 \), so \( T_1=372 \) K and \( T_2=310 \) K.

Teacher's Note:
1. Write two equations from the two efficiencies.
2. Check: \( 1-\dfrac{310}{372}=\dfrac16 \).

Answer: (d) \( \dfrac p2,T \)

Explanation:
1. Free expansion into vacuum in an insulated container: \( Q=0 \) and \( W=0 \), so \( \Delta U=0 \).
2. For an ideal gas the temperature is unchanged: \( T \).
3. The volume doubles, so \( P=\dfrac p2 \).

Teacher's Note:
1. Free expansion does no work, and an ideal gas has no intermolecular forces.
2. The temperature of a real gas would drop slightly.

Answer: (d) \( \dfrac1{2\pi}\sqrt{\dfrac{A^{2}\gamma P_0}{MV_0}} \)

Explanation:
1. An isolated system means adiabatic: \( PV^{\gamma}=\) const, so \( dP=-\gamma\dfrac{P_0}{V_0}dV \).
2. For a displacement \( x \), \( dV=Ax \) and the restoring force is \( A\,dP=-\dfrac{\gamma P_0A^{2}}{V_0}x \).
3. \( \omega^{2}=\dfrac{\gamma P_0A^{2}}{MV_0} \), so \( f=\dfrac1{2\pi}\sqrt{\dfrac{A^{2}\gamma P_0}{MV_0}} \).

Teacher's Note:
1. The effective spring constant of a gas is \( \gamma P_0A^{2}/V_0 \).
2. Adiabatic oscillation uses \( \gamma P \), isothermal would use \( P \).

Answer: (c) \( \dfrac{9P_0V_0}{4nR} \)

Explanation:
1. Line: \( P=3P_0-\dfrac{P_0}{V_0}V \).
2. \( PV=3P_0V-\dfrac{P_0V^{2}}{V_0} \) is maximum at \( V=\dfrac32V_0 \), giving \( PV=\dfrac94P_0V_0 \).
3. \( T_{max}=\dfrac{9P_0V_0}{4nR} \).

Teacher's Note:
1. Find where \( PV \) is largest, not the end points.
2. The vertex is at half of the intercept volume \( 3V_0 \).

Answer: (a) \( n=\dfrac{C-C_p}{C-C_v} \)

Explanation:
1. For \( PV^{n}= \) const the molar heat capacity is \( C=C_v+\dfrac{R}{1-n} \).
2. With \( R=C_p-C_v \): \( 1-n=\dfrac{C_p-C_v}{C-C_v} \).
3. \( n=1-\dfrac{C_p-C_v}{C-C_v}=\dfrac{C-C_p}{C-C_v} \).

Teacher's Note:
1. Check: \( C=0 \) gives \( n=\dfrac{C_p}{C_v}=\gamma \) (adiabatic), \( C=\infty \) gives \( n=1 \).
2. Both limits agree with known processes.

Answer: (c) \( a=14b \)

Explanation:
1. Specific heats per unit mass: \( C_p-C_v=\dfrac RM \).
2. \( a=\dfrac R2 \) for hydrogen and \( b=\dfrac R{28} \) for nitrogen.
3. \( \dfrac ab=\dfrac{28}{2}=14 \), so \( a=14b \).

Teacher's Note:
1. The difference of specific heats per unit mass is inversely proportional to the molar mass.
2. Hydrogen is 14 times lighter than nitrogen per mole.

Answer: (a) (1) 189 K (2) -2.77 kJ

Explanation:
1. Adiabatic: \( TV^{\gamma-1}=\) const with \( \gamma=\tfrac53 \).
2. \( T_2=300\times2^{-2/3}\approx189 \) K.
3. \( \Delta U=nC_v\Delta T=2\times1.5\times8.314\times(189-300)\approx-2.77 \) kJ.

Teacher's Note:
1. An adiabatic expansion cools the gas, so \( \Delta U \) is negative.
2. Use \( \gamma=\tfrac53 \) for a monatomic gas.

Answer: (c) \( \dfrac3T \)

Explanation:
1. \( P\propto\dfrac TV \) so \( PT^{2}\propto\dfrac{T^{3}}{V}=\) const, giving \( V\propto T^{3} \).
2. Coefficient of volume expansion \( \gamma=\dfrac1V\dfrac{dV}{dT}=\dfrac3T \).

Teacher's Note:
1. Eliminate \( P \) with \( PV=nRT \) to relate \( V \) and \( T \).
2. For \( V\propto T^{n} \) the coefficient is \( n/T \).

Answer: (c) \( \dfrac75 \)

Explanation:
1. \( C_v=C_p-R=\tfrac72R-R=\tfrac52R \).
2. \( \gamma=\dfrac{C_p}{C_v}=\dfrac{7/2}{5/2}=\dfrac75 \).

Teacher's Note:
1. This is the value for a rigid diatomic gas.
2. Always subtract \( R \) to get \( C_v \).

Answer: (c) monoatomic

Explanation:
1. \( C_v=\dfrac{R}{0.67}\approx1.5R \).
2. \( C_v=\tfrac32R \) corresponds to three degrees of freedom, a monoatomic gas.

Teacher's Note:
1. Match \( C_v \) to \( \tfrac f2R \).
2. \( \gamma=\dfrac53 \) for monoatomic gases.

Answer: (a) \( 120R \)

Explanation:
1. \( P=\dfrac{RT}{V}=\dfrac{R}{kT} \) and \( dV=2kT\,dT \).
2. \( W=\int P\,dV=\int\dfrac{R}{kT}2kT\,dT=2R\,\Delta T \).
3. \( W=2R\times60=120R \).

Teacher's Note:
1. In general for \( V\propto T^{n} \), \( W=nR\Delta T \).
2. Substitute \( P \) in terms of \( T \) first.

Answer: (b) \( 3R \)

Explanation:
1. \( \Delta U=Q-W=Q-\dfrac Q2=\dfrac Q2 \).
2. \( \Delta U=nC_v\Delta T \) so \( nC\Delta T=Q=2\Delta U=2nC_v\Delta T \).
3. \( C=2C_v=2\times\tfrac32R=3R \).

Teacher's Note:
1. Half of the heat goes into internal energy, so \( C=2C_v \).
2. Use \( C_v=\tfrac32R \) for monatomic gases.

Answer: (c) 1600 J

Explanation:
1. \( \eta=1-\dfrac{300}{600}=\dfrac12 \).
2. \( \eta=\dfrac{W}{Q_1} \), so \( Q_1=\dfrac{800}{0.5}=1600 \) J.

Teacher's Note:
1. Efficiency of a Carnot engine depends only on the reservoir temperatures.
2. Heat supplied is work divided by efficiency.

Answer: (d) 460 J

Explanation:
1. Along ABC: \( W_{BC}=6\times10^{4}\times2\times10^{-3}=120 \) J and \( W_{AB}=0 \).
2. \( Q=400+100=500 \) J, so \( \Delta U_{AC}=500-120=380 \) J.
3. Along the straight line AC: \( W=\tfrac12(2\times10^{4}+6\times10^{4})(2\times10^{-3})=80 \) J, so \( Q_{AC}=380+80=460 \) J.

Teacher's Note:
1. \( \Delta U \) is the same for ABC and AC.
2. Work is the area under each path on the P-V diagram.

Answer: (c) 8 atm

Explanation:
1. Sudden compression is adiabatic: \( P_2=P_1\left(\dfrac{V_1}{V_2}\right)^{\gamma}=1\times4^{3/2} \).
2. \( 4^{3/2}=8 \), so the final pressure is 8 atm.

Teacher's Note:
1. \( 4^{3/2}=\left(\sqrt4\right)^{3} \).
2. NTP pressure is 1 atm.

Answer: (a) \( \dfrac{P}{n} \)

Explanation:
1. Taking logarithms, \( \ln V+n\ln P= \) const, so \( \dfrac{dV}{V}+n\dfrac{dP}{P}=0 \).
2. \( B=-V\dfrac{dP}{dV}=-\dfrac{dP}{dV/V}=\dfrac{P}{n} \).

Teacher's Note:
1. Differentiate the logarithm of the process equation.
2. Check \( n=1 \): \( VP= \) const (isothermal), \( B=P \).

Answer: (c) \( -\gamma \)

Explanation:
1. Zero molar heat capacity means \( Q=0 \) throughout, which is an adiabatic process.
2. For an adiabatic process \( pV^{\gamma}= \) constant, that is \( p=aV^{-\gamma} \).
3. Comparing with \( p=aV^{b} \): \( b=-\gamma \).

Teacher's Note:
1. The pressure falls as the volume grows, so the exponent must be negative.
2. Compare with the adiabatic law directly.