JEE Physics Thermodynamics MCQs Set 16

Answer: (d) 13 kJ

Explanation:
1. \( \Delta U=Q-W=16-20=-4 \) kJ (it depends only on the end states).
2. Second path: \( W=Q-\Delta U=9-(-4)=13 \) kJ.

Teacher's Note:
1. Internal energy is a state function, so \( \Delta U \) is the same for both paths.
2. Work and heat are path dependent.

Answer: (a) 680 joule

Explanation:
1. \( Q=400\times4.2=1680 \) J.
2. \( \Delta U=Q-W=1680-1000=680 \) J.

Teacher's Note:
1. Work done by the system is positive in \( \Delta U=Q-W \).
2. Convert calories to joules first.

Answer: (d) -60 calorie

Explanation:
1. \( \Delta U=Q-W=-50-(-20)=-30 \) cal.
2. Final \( U=-30+(-30)=-60 \) cal.

Teacher's Note:
1. Heat leaving and work done on the system both have negative signs here.
2. Final energy equals initial energy plus the change.

Answer: (a) \( \dfrac13R(T_2-T_1) \)

Explanation:
1. \( P=aV^{2} \), so \( W=\int P\,dV=\dfrac a3\left(V_2^{3}-V_1^{3}\right) \).
2. \( PV=RT \) gives \( aV^{3}=RT \), so \( W=\dfrac13R(T_2-T_1) \).

Teacher's Note:
1. Use the equation of state to convert \( PV \) into \( RT \).
2. For \( P\propto V^{n} \), \( W=\dfrac{R\Delta T}{n+1} \).

Answer: (d) 1 : 2 : 2

Explanation:
1. \( T\propto PV \): \( T_1=P_0V_0 \), \( T_2=4P_0V_0 \), \( T_3=P_0(4V_0)=4P_0V_0 \).
2. Mean speed \( \propto\sqrt T \): \( 1:\sqrt4:\sqrt4=1:2:2 \).

Teacher's Note:
1. Use \( PV=nRT \) to find the temperature in each state.
2. Speed goes as the square root of temperature.

Answer: (d) 24 J

Explanation:
1. Along acb: \( \Delta U=Q-W=200-80=120 \) J.
2. Along adb the same \( \Delta U \) applies: \( W=Q-\Delta U=144-120=24 \) J.

Teacher's Note:
1. \( \Delta U \) depends only on the end states a and b.
2. Use the first law on the second path.

Answer: (a) -172 J

Explanation:
1. Going back, \( \Delta U_{ba}=-120 \) J.
2. Work done on the system means \( W=-52 \) J.
3. \( Q=\Delta U+W=-120-52=-172 \) J (heat is given out).

Teacher's Note:
1. Reversing the path reverses the sign of \( \Delta U \).
2. Work done on the system is negative in \( Q=\Delta U+W \).

Answer: (b) \( \dfrac{60}{R} \)

Explanation:
1. Kinetic energy per mole \( =\tfrac12Mv^{2}=\tfrac12(0.03)(100)^{2}=150 \) J.
2. This raises the internal energy: \( 150=C_v\Delta T=\tfrac52R\,\Delta T \).
3. \( \Delta T=\dfrac{60}{R} \).

Teacher's Note:
1. Bulk kinetic energy is converted into random thermal energy.
2. Use the molar mass in kg/mol and \( C_v=\tfrac52R \) for a diatomic gas.

Answer: (a) \( \gamma:\gamma-1 \)

Explanation:
1. \( V\propto T \) means the pressure is constant (isobaric).
2. \( \Delta Q=nC_p\Delta T \) and \( \Delta W=nR\Delta T \).
3. \( \dfrac{\Delta Q}{\Delta W}=\dfrac{C_p}{C_p-C_v}=\dfrac{\gamma}{\gamma-1} \).

Teacher's Note:
1. A straight line through the origin on a V-T graph is an isobar.
2. \( C_p-C_v=R \).

Answer: (d) 12600 J

Explanation:
1. At constant volume \( Q=\Delta U=nC_v\Delta T \): \( nC_v=\dfrac{6300}{150}=42 \) J/K.
2. Internal energy depends only on temperature, so for 300 K: \( \Delta U=42\times300=12600 \) J.

Teacher's Note:
1. \( \Delta U=nC_v\Delta T \) for any process of an ideal gas.
2. The constant-pressure data is not needed here.

Answer: (d) \( -2.0\times10^{7}\ \text{N m}^{-5} \)

Explanation:
1. Adiabatic: \( PV^{\gamma}=\) constant, so \( \dfrac{dP}{dV}=-\gamma\dfrac PV \).
2. \( =-1.4\times\dfrac{0.7\times10^{5}}{0.0049}=-2.0\times10^{7}\ \text{N m}^{-5} \).

Teacher's Note:
1. An adiabat is steeper than an isotherm by the factor \( \gamma \).
2. The slope is negative on a P-V diagram.

Answer: (b) \( P_1d_1^{-\gamma}=P_2d_2^{-\gamma} \)

Explanation:
1. \( PV^{\gamma}=\) constant and \( V\propto\dfrac1d \), so \( Pd^{-\gamma}=\) constant.
2. \( P_1d_1^{-\gamma}=P_2d_2^{-\gamma} \), equivalently \( \dfrac{P_1}{P_2}=\left(\dfrac{d_1}{d_2}\right)^{\gamma} \).

Teacher's Note:
1. Replace volume by 1/density in the adiabatic law.
2. Check the sign of the exponent: pressure rises with density.

Answer: (c) \( P^{1-\gamma}T^{\gamma} \)

Explanation:
1. \( PV^{\gamma}=\) const and \( V=\dfrac{nRT}{P} \) give \( P\left(\dfrac TP\right)^{\gamma}=\) const.
2. \( P^{1-\gamma}T^{\gamma}=\) constant.

Teacher's Note:
1. Eliminate \( V \) with the ideal gas equation.
2. Check \( \gamma=1 \): \( T \) is constant, an isothermal change.

Answer: (b) 0.25

Explanation:
1. Heat rejected \( Q_2=3W \).
2. Heat taken \( Q_1=W+Q_2=4W \).
3. \( \eta=\dfrac WQ_1=\dfrac14=0.25 \).

Teacher's Note:
1. Heat in equals work out plus heat rejected.
2. Efficiency is work divided by heat absorbed.

Answer: (c) 253 K

Explanation:
1. \( \eta=1-\dfrac{273}{373}=\dfrac{100}{373}\approx0.268 \).
2. New efficiency \( =1.2\times0.268\approx0.322 \).
3. \( 1-\dfrac{T_2}{373}=0.322 \), so \( T_2\approx253 \) K.

Teacher's Note:
1. Take the temperatures in kelvin.
2. A '20% increase' multiplies the efficiency by 1.2.

Answer: (d) \( 300\sqrt2 \)

Explanation:
1. With \( \gamma=\dfrac43 \), \( T\propto P^{(\gamma-1)/\gamma}=P^{1/4} \).
2. \( T_2=300\times4^{1/4}=300\sqrt2 \) K.

Teacher's Note:
1. Use \( P^{1-\gamma}T^{\gamma}=\) constant.
2. The exponent \( \dfrac{\gamma-1}{\gamma} \) is \( \dfrac14 \) here.

Answer: (b) \( 2RT_0 \)

Explanation:
1. \( P=aV \) gives \( W=\int aV\,dV=\dfrac a2(V_2^{2}-V_1^{2})=\dfrac12R(T_2-T_1)=\dfrac12RT_0 \) (using \( aV^{2}=RT \)).
2. \( \Delta U=\tfrac32R(T_2-T_1)=\tfrac32RT_0 \).
3. \( Q=\Delta U+W=2RT_0 \).

Teacher's Note:
1. Use \( PV=RT \) with \( P\propto V \) to get \( W \) from the temperatures.
2. Monatomic gas: \( C_v=\tfrac32R \).

Answer: (d) \( \dfrac57 \)

Explanation:
1. At constant pressure \( Q=nC_p\Delta T \) and \( \Delta U=nC_v\Delta T \).
2. Fraction \( =\dfrac{C_v}{C_p}=\dfrac1\gamma=\dfrac57 \) for a diatomic gas.

Teacher's Note:
1. The rest of the heat, \( \tfrac27 \), is the expansion work.
2. \( \gamma=\tfrac75 \) for a rigid diatomic gas.

Answer: (d) \( \Delta U \) is greatest under adiabatic process

Explanation:
1. In an expansion the adiabatic process cools the gas (\( \Delta U<0 \)), the isothermal one has \( \Delta U=0 \) and the isobaric one heats it (\( \Delta U>0 \)).
2. So \( \Delta U \) is least for the adiabatic process and greatest for the isobaric process, with the isothermal value in between.
3. The statement that \( \Delta U \) is greatest under the adiabatic process is false.

Teacher's Note:
1. Compare the end temperatures reached from the same initial state.
2. Isobaric expansion needs heat and raises the temperature.

Answer: (a) -10 calorie

Explanation:
1. Internal energy is a state function, so \( \Delta U \) depends only on the end states.
2. \( \Delta U_{a\to b}=10 \) cal, hence \( \Delta U_{b\to a}=-10 \) cal along any path.

Teacher's Note:
1. Reversing the direction reverses the sign of \( \Delta U \).
2. The shape of the path is irrelevant for \( \Delta U \).

Answer: (b) -40 J

Explanation:
1. During 2 → 3, \( Q=\Delta U=nC_v(T_3-T_2)=nC_v(275-300)=-40 \) J, so \( nC_v=1.6 \) J/K.
2. Adiabatic 3 → 1: \( W=-\Delta U=-nC_v(T_1-T_3)=-1.6\times25 \).
3. \( W=-40 \) J (work is done on the gas).

Teacher's Note:
1. In an adiabatic compression the work done on the gas equals the increase in internal energy.
2. Constant volume means no work in process 2 → 3.

Answer: (c) It is not applicable to any cyclic process

Explanation:
1. The first law is a statement of conservation of energy and defines the internal energy.
2. It applies to all processes, including cyclic ones (where \( \Delta U=0 \) and \( Q=W \)).
3. The concept of entropy is introduced by the second law and not the first, but the claim in (a) is the plainly false one: the first law holds for every process.

Teacher's Note:
1. A cyclic process is a standard application of the first law.
2. Entropy belongs to the second law.