JEE Physics Thermodynamics MCQs Set 07

Answer: (c) \(3.5R\)

Explanation:
1. With \(PV = RT\), \(PT = RT^2/V = \text{const}\), so \(V \propto T^2\) or \(V = kT^2\).
2. \(dV = 2kT\,dT\) and \(P\,dV = (RT/V)\,2kT\,dT = 2R\,dT\).
3. \(dQ = C_v dT + P\,dV = (1.5R + 2R)\,dT\), so \(C = 3.5R\).

Teacher's Note:
1. For any process \(V \propto T^m\), the molar heat capacity is \(C_v + R/m\).

Answer: (b) \(\tfrac{3}{2}P_0V_0\)

Explanation:
1. \(PT = \text{const}\), so when \(P\) halves, \(T\) doubles.
2. Initially \(P_0V_0 = 2RT_0\), so \(T_0 = P_0V_0/2R\).
3. \(\Delta U = nC_v\Delta T = 2 \times \tfrac{3}{2}R \times T_0 = 3R \times \dfrac{P_0V_0}{2R} = \tfrac{3}{2}P_0V_0\).

Teacher's Note:
1. The change in internal energy depends only on the temperature change.

Answer: (a) an isobaric process

Explanation:
1. Work is the area under the P-V curve.
2. From the same initial point, the isobaric line stays highest, the isotherm falls below it and the adiabat falls further.
3. So the isobaric process gives the largest work.

Teacher's Note:
1. Pressure is highest throughout the isobaric expansion, and falls fastest in an adiabatic one.

Answer: (c) change in internal energy \(= 0\)

Explanation:
1. \(T_A = P_0V_0/R = T_0\).
2. At C, \(P_CV_C = (P_0/2)(2V_0) = P_0V_0\), so \(T_C = T_0\) and \(\Delta U = 0\).
3. Work done \(= P_0 \times V_0 = RT_0\) (only along AB), so heat absorbed is \(RT_0\).
4. So only option c is correct.

Teacher's Note:
1. Equal products \(PV\) at the two ends mean equal temperatures, so \(\Delta U = 0\).

Answer: (d) 900 K

Explanation:
1. For an adiabatic process \(TV^{\gamma-1} = \text{const}\).
2. \(T_2 = T_1(V_1/V_2)^{0.5} = 300 \times 9^{0.5} = 900\) K.

Teacher's Note:
1. Convert to kelvin before applying the formula.

Answer: (b) 1.50

Explanation:
1. \(C_v\) of the mixture \(= (\tfrac{3}{2}R + \tfrac{5}{2}R)/2 = 2R\).
2. \(C_p = C_v + R = 3R\).
3. \(\gamma = 3R/2R = 1.5\).

Teacher's Note:
1. Average the molar heat capacities, then compute \(\gamma\). Do not average the values of \(\gamma\).

Answer: (b) \(-4\) J

Explanation:
1. Along the isotherm a to b, \(\Delta U = 0\).
2. Along the adiabatic b to c, \(Q = 0\), so \(\Delta U = -W = -4\) J.
3. Internal energy is a state function, so the change from a to c by any path is \(0 - 4 = -4\) J.

Teacher's Note:
1. Internal energy depends only on end states, so the shape of the path does not matter.

Answer: (a) The maximum pressure during A-B-C-D is at D.

Explanation:
1. A to B lies on a line through the origin, so the pressure stays \(P_0\).
2. B to C is isothermal compression, so the pressure doubles to \(2P_0\).
3. C to D is isochoric at \(V_0\) with temperature doubled, so the pressure doubles again to \(4P_0\), the maximum, at D.
4. Work from A to B is \(P_0V_0\) but B to C gives \(-2P_0V_0 \ln 2\), which is larger in magnitude, so the net work is negative.

Teacher's Note:
1. On a V-T diagram, a line through the origin is an isobar. Every point on it has the same pressure.

Answer: (c) 30 J

Explanation:
1. For an ideal gas \(\Delta U = \tfrac{3}{2}\Delta(PV)\) for a monoatomic gas.
2. \(P_AV_A = 100\times10^3 \times 100\times10^{-6} = 10\) J and \(P_CV_C = 200\times10^3 \times 150\times10^{-6} = 30\) J.
3. \(\Delta U = 1.5 \times (30 - 10) = 30\) J.

Teacher's Note:
1. The temperature given for B is not needed once you work with the products \(PV\).

Answer: (c) CD

Explanation:
1. In an isochoric process \(Q = nC_v\Delta T = \dfrac{C_v}{R}\,V\,\Delta P\).
2. The pressure fall is the same for both, so the heat exchanged is proportional to \(V\).
3. CD is at larger volume, so it exchanges more heat.

Teacher's Note:
1. For a given pressure change at constant volume, a larger volume means a larger temperature change.

Answer: (a) 62.1 J

Explanation:
1. \(C_p = 207/10 = 20.7\) J mol\(^{-1}\)K\(^{-1}\).
2. \(C_v = C_p - R = 20.7 - 8.3 = 12.4\) J mol\(^{-1}\)K\(^{-1}\).
3. \(Q = 12.4 \times 5 = 62\) J, which is closest to 62.1 J.

Teacher's Note:
1. Mayer's relation \(C_p - C_v = R\) links the two heat capacities.

Answer: (c) \(-15\) J

Explanation:
1. Work in A to B is \(P\Delta V = 20 \times 1 = 20\) J.
2. Work in B to C is zero because the volume is constant.
3. For a cycle \(\Delta U = 0\), so net work equals net heat, \(5 = 20 + 0 + W_{CA}\), giving \(W_{CA} = -15\) J.

Teacher's Note:
1. Over a full cycle the net work done equals the net heat absorbed.

Answer: (d) 13 kJ

Explanation:
1. First path: \(\Delta U = Q - W = 16 - 20 = -4\) kJ.
2. The same \(\Delta U\) applies to the second path.
3. \(W = Q - \Delta U = 9 - (-4) = 13\) kJ.

Teacher's Note:
1. \(\Delta U\) is path independent, while \(Q\) and \(W\) separately depend on the path.

Answer: (b) \(20R\)

Explanation:
1. \(W = \int P\,dV\) with \(P = RT/V\).
2. \(dV = \tfrac{2}{3}kT^{-1/3}dT\) and \(V = kT^{2/3}\), so \(P\,dV = \tfrac{2}{3}R\,dT\).
3. \(W = \tfrac{2}{3}R \times 30 = 20R\).

Teacher's Note:
1. For \(V \propto T^m\), the work is \(mR\,\Delta T\).

Answer: (a) A to B isochoric with pressure rising, B to C isothermal expansion with pressure falling, C to A isobaric compression

Explanation:
1. A to B is at constant volume with rising \(T\), so \(P\) rises.
2. B to C is at constant \(T\) with rising \(V\), so \(P\) falls along a hyperbola.
3. C to A lies on a line through the origin of the V-T plot, which is an isobar, so the gas is compressed at constant pressure back to A.
4. Only option a matches.

Teacher's Note:
1. In a V-T plot a horizontal line is isochoric, a vertical line is isothermal and a line through the origin is isobaric.

Answer: (b) \(T_0/4\)

Explanation:
1. After isothermal expansion the state is \((P_0/2,\; 2V_0,\; T_0)\).
2. Compression at constant pressure \(P_0/2\) to \(V_0/2\) gives \(T \propto V\), so \(T = T_0 \times \dfrac{V_0/2}{2V_0} = T_0/4\).
3. Check: the final state \((P_0/2, V_0/2)\) lies on the line \(P \propto V\) through \((P_0, V_0)\).

Teacher's Note:
1. At constant pressure the temperature is proportional to volume (Charles' law).

Answer: (b) \(2\sqrt{2} : 1 : 2\)

Explanation:
1. Adiabatic: \(P_AV^{\gamma} = P\,(2V)^{\gamma}\), so \(P_A = 2^{3/2}P = 2\sqrt{2}P\).
2. Isobaric: \(P_B = P\).
3. Isothermal: \(P_C V = P \cdot 2V\), so \(P_C = 2P\).
4. The ratio is \(2\sqrt{2} : 1 : 2\).

Teacher's Note:
1. Write the process law for each sample and solve for the initial pressure in terms of the common final pressure.