JEE Physics Thermodynamics MCQs Set 09

Answer: (a) greater than the slope of the isothermal curve

Explanation:
1. Isothermal: \( \left(\frac{dP}{dV}\right)_T=-\frac PV \).
2. Adiabatic: \( \left(\frac{dP}{dV}\right)_{ad}=-\gamma\frac PV \).
3. Since \( \gamma>1 \), the adiabatic curve is steeper at every common point.

Teacher's Note:
1. The magnitude of the slope of an adiabat is γ times that of the isotherm through the same point.
2. This is why an adiabat falls faster than an isotherm.

Answer: (b) \( C_v(T_1-T_2) \)

Explanation:
1. Adiabatic: \( Q=0\Rightarrow W=-\Delta U \).
2. \( W=-nC_v(T_2-T_1)=C_v(T_1-T_2) \) for one mole.

Teacher's Note:
1. Work in an adiabatic expansion comes entirely from the internal energy, so the gas cools.
2. Also W = R(T1 − T2)/(γ − 1) = Cv(T1 − T2).

Answer: (b) Internal energy will remain constant

Explanation:
1. The internal energy of an ideal gas depends only on temperature.
2. Temperature is constant, so \( \Delta U=0 \) and the heat absorbed equals the work done.

Teacher's Note:
1. In an isothermal change Q = W.
2. Real gases may differ because of intermolecular forces, but the ideal gas has none.

Answer: (b) 60 cal

Explanation:
1. \( W=P\Delta V=nR\Delta T \) and at constant pressure doubling V doubles T, so \( \Delta T=300\ \text{K} \).
2. \( W=0.1\times2\times300=60\ \text{cal} \)

Teacher's Note:
1. Convert the temperature to kelvin: 27 °C = 300 K.
2. At constant pressure V ∝ T.

Answer: (d) \( p^{1-\gamma}T^{\gamma}=\text{constant} \)

Explanation:
1. \( pV^\gamma=\text{const} \) and \( V=\frac{nRT}{p} \).
2. \( p\left(\frac{T}{p}\right)^\gamma=\text{const}\Rightarrow p^{1-\gamma}T^\gamma=\text{const} \)

Teacher's Note:
1. Eliminate V between the adiabatic relation and the ideal gas equation.
2. Equivalent form: T^γ / p^(γ−1) = constant.

Answer: (d) 891 °C

Explanation:
1. For a monatomic gas \( \gamma=\frac53 \) and \( T_1=18+273=291\ \text{K} \).
2. \( T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=291\times8^{2/3}=291\times4=1164\ \text{K} \).
3. \( 1164-273=891\ ^\circ\text{C} \)

Teacher's Note:
1. Use TV^(γ−1) = constant and convert to kelvin before applying it.
2. 8^(2/3) = 4.

Answer: (b) 1.2

Explanation:
1. \( T_1=500\ \text{K},\ T_2=400\ \text{K} \).
2. \( \eta=1-\frac{400}{500}=0.2 \).
3. \( W=\eta Q_1=0.2\times6=1.2\ \text{kcal} \)

Teacher's Note:
1. Temperatures must be in kelvin for the Carnot efficiency.
2. Work = efficiency × heat absorbed.

Answer: (c) Heat given to a system is used up in raising the temperature only

Explanation:
1. At constant pressure \( V\propto T \), \( W=P\Delta V=\mu R\Delta T \) are correct.
2. Heat supplied is \( Q=\mu C_p\Delta T=\Delta U+W \): part raises the temperature (internal energy) and part does work, so statement (c) is incorrect.

Teacher's Note:
1. In an isobaric process the gas expands as it is heated, so it does work.
2. Only in an isochoric process is the heat used entirely to raise the temperature.

Answer: (c) The work done W is always greater than zero

Explanation:
1. In a cycle the initial and final states coincide and \( \Delta U=0 \), so \( Q=W \) (statements 1 and 2 are correct).
2. The net work depends on the direction: it is positive for a clockwise cycle on the P-V diagram (heat engine) but negative for an anticlockwise one (refrigerator), so W is not always positive.

Teacher's Note:
1. The sign of the net work depends on whether the loop is traversed clockwise or anticlockwise.
2. Net work is the area enclosed by the loop, with a sign.

Answer: (c) isochoric

Explanation:
1. For an ideal gas \( \frac PT=\frac{nR}{V} \).
2. \( V=k\left(\frac{nR}{V}\right)^{0.33}\Rightarrow V^{1.33}=k(nR)^{0.33} \).
3. The right side is a constant, so V is constant: the process is isochoric.

Teacher's Note:
1. Replace P/T by nR/V and solve for V.
2. A relation that gives a constant V means an isochoric process.

Answer: (d) \( \left(\frac\pi2+4\right)\text{atm L} \)

Explanation:
1. Work = area under the curve between V = 2 L and V = 4 L.
2. Rectangle below P = 2 atm: \( 2\times2=4\ \text{atm L} \).
3. Half of the circle (radius 1 in both units) above it: \( \frac12\pi(1)(1)=\frac\pi2\ \text{atm L} \).
4. Total: \( \left(4+\frac\pi2\right)\text{atm L} \)

Teacher's Note:
1. Work done by a gas is the area under the P-V curve down to the volume axis.
2. Split the area into a rectangle and a half-ellipse.

Answer: (c) \( Q_1>Q_2 \)

Explanation:
1. Both processes have the same initial and final states, so \( \Delta U \) is the same.
2. Work done is the area under the curve, which is larger for the upper path I: \( W_1>W_2 \).
3. \( Q=\Delta U+W \), so \( Q_1>Q_2 \)

Teacher's Note:
1. Internal energy is a state function, but heat and work depend on the path.
2. A bigger area under the curve means more work, hence more heat.

Answer: (d) 40000 J

Explanation:
1. Work is done only in the constant-pressure steps BC and DA (AB and CD are at constant volume).
2. \( W_{BC}=nR(T_C-T_B)=6R(2200-800)=8400R \).
3. \( W_{DA}=nR(T_A-T_D)=6R(600-1200)=-3600R \).
4. \( W=4800R=4800\times8.31\approx4\times10^4\ \text{J} \)

Teacher's Note:
1. Lines through the origin on a P-T diagram are isochoric.
2. Net work is the sum of the isobaric works, with sign.

Answer: (d) all of the above statements are correct

Explanation:
1. Internal energy is a state function, so the change depends only on the end states and not on the path.
2. In a cyclic process the end states coincide, so the change is zero.

Teacher's Note:
1. State functions: U, P, V, T. Path functions: Q, W.
2. All three given statements describe the same property.

Answer: (c) negative, zero, positive

Explanation:
1. A to B: the volume decreases (compression), so the work done by the gas is negative.
2. B to C: the volume is constant, so the work is zero.
3. C to A: the volume increases at constant pressure (expansion), so the work is positive.

Teacher's Note:
1. The sign of W follows the sign of ΔV.
2. Work is zero whenever the volume stays constant.

Answer: (c) heat is taken by the system

Explanation:
1. The volume increases, so work is done by the system (not on it).
2. Along the line, \( P=\frac{P_0}{V_0}V \) gives \( PV=\frac{P_0}{V_0}V^2 \), which increases, so the temperature rises: \( \Delta U>0 \).
3. \( Q=\Delta U+W>0 \), so heat is absorbed.

Teacher's Note:
1. PV ∝ T, so compare PV at the two states: P0V0 → 4P0V0 means T quadruples.
2. Both ΔU and W are positive, so Q is positive.

Answer: (b) Heat is a path function

Explanation:
1. Heat and work depend on the process followed (path functions).
2. Internal energy depends only on the state (state function).

Teacher's Note:
1. State functions have exact differentials; Q and W do not.
2. The first law links the path functions Q and W with the state function U.

Answer: (a) A graph of PV against V: a horizontal straight line

Explanation:
1. Isothermal: \( PV=nRT=\text{constant} \).
2. So a plot of PV against V is a horizontal straight line, while P against V is a hyperbola.

Teacher's Note:
1. Check each graph against PV = constant.
2. A horizontal V-P line would mean constant volume, not constant temperature.

Answer: (a) 40%

Explanation:
1. \( \frac WQ=\frac{R\Delta T}{C_p\Delta T}=\frac R{C_p}=\frac{\gamma-1}{\gamma} \).
2. \( =\frac{2/3}{5/3}=\frac25=40\% \)

Teacher's Note:
1. For a monatomic gas Cp = 5R/2, so the fraction is 2/5.
2. The remaining 60% raises the internal energy.

Answer: (a) \( P_1>P_2 \)

Explanation:
1. \( V=\frac{nR}{P}T \), so the slope of the V-T line is \( \frac{nR}{P} \).
2. A steeper line (P2) means a smaller pressure, so \( P_1>P_2 \)

Teacher's Note:
1. Slope of V-T is inversely proportional to pressure.
2. Compare lines at the same temperature: P2 gives the larger volume.

Answer: (d) 2P

Explanation:
1. Isothermal: \( P\cdot V=P_1\cdot4V\Rightarrow P_1=\frac P4 \).
2. Adiabatic: \( P_1(4V)^{1.5}=P_2V^{1.5}\Rightarrow P_2=\frac P4\times4^{1.5}=\frac P4\times8=2P \)

Teacher's Note:
1. Use PV = constant for the isothermal step and PV^γ = constant for the adiabatic step.
2. 4^1.5 = 8.

Answer: (b) W

Explanation:
1. \( \Delta Q=\Delta U+W=0+W=W \)

Teacher's Note:
1. All the heat absorbed is converted into work.
2. Temperature stays constant, so internal energy is unchanged.

Answer: (a) 3.5 atmosphere

Explanation:
1. Both gases occupy the same volume V at the same temperature after mixing, and each exerts its original pressure (partial pressure) because each previously filled the same volume V at that temperature.
2. Dalton's law: \( P=2.5+1=3.5\ \text{atm} \)

Teacher's Note:
1. Partial pressure of each gas is determined by n, T and V; none of them changed.
2. Total pressure is the sum of the partial pressures.

Answer: (b) 5/2

Explanation:
1. For an adiabatic change \( P^{1-\gamma}T^\gamma=\text{const}\Rightarrow P\propto T^{\gamma/(\gamma-1)} \).
2. For a monatomic gas \( \gamma=\frac53 \), so \( C=\frac{5/3}{2/3}=\frac52 \)

Teacher's Note:
1. Use the pressure-temperature form of the adiabatic relation.
2. For a diatomic gas the same formula gives 7/2.

Answer: (c) \( \Delta U_{ab}=\Delta U_{ac} \) but \( Q_{ac}>Q_{ab} \)

Explanation:
1. b and c lie on the same isotherm T2, so the internal energy of the ideal gas is the same there: \( \Delta U_{ab}=\Delta U_{ac} \).
2. a to b is isochoric: \( W_{ab}=0 \). a to c is an isobaric expansion: \( W_{ac}>0 \).
3. \( Q=\Delta U+W \), so \( Q_{ac}>Q_{ab} \)

Teacher's Note:
1. ΔU depends only on the temperatures at the end points for an ideal gas.
2. The isobaric path does extra work, so it needs more heat.

Answer: (c) equal in A, B and C

Explanation:
1. Internal energy is a state function, so the change depends only on the initial and final states.
2. All three paths join the same two states, so \( \Delta U_A=\Delta U_B=\Delta U_C \)

Teacher's Note:
1. Only heat and work depend on the path.
2. Work is largest on the topmost path, A.

Answer: (c) \( \frac\pi4(P_2-P_1)(V_2-V_1) \)

Explanation:
1. The work in a cycle is the enclosed area, here an ellipse with semi-axes \( \frac{P_2-P_1}{2} \) and \( \frac{V_2-V_1}{2} \).
2. \( \text{Area}=\pi ab=\frac\pi4(P_2-P_1)(V_2-V_1) \)

Teacher's Note:
1. Area of an ellipse is π times the product of its semi-axes.
2. Dimensional check: pressure × volume = energy.

Answer: (b) \( U_B<U_A \)

Explanation:
1. The internal energy of an ideal gas depends only on temperature.
2. TA = TD = T and TB = TC = 2T, so UA = UD and UB = UC; also UC > UD since 2T > T.
3. Hence UB < UA is the false statement, because UB corresponds to 2T which is greater than T.

Teacher's Note:
1. Read the temperatures of each state from the horizontal axis.
2. Only the temperature matters for U of an ideal gas.

Answer: (a) 50 J

Explanation:
1. Adiabatic: \( Q=0 \). First law: \( \Delta U=Q-W_{by}=-W_{by} \).
2. Work done on the gas is 50 J, so \( W_{by}=-50\ \text{J} \) and \( \Delta U=+50\ \text{J} \)

Teacher's Note:
1. Work done on the gas increases its internal energy when no heat flows.
2. The number of moles is not needed.