JEE Physics Thermodynamics MCQs Set 04

Answer: (d) the law of conservation of energy

Explanation:
1. The first law, \( Q = \Delta U + W \), states that heat supplied is accounted for by the rise in internal energy and the work done.
2. It is a statement of conservation of total energy, including heat.

Teacher's Note:
1. Mechanical energy alone is not conserved when heat is involved.

Answer: (c) 2193 joules

Explanation:
1. Heat absorbed \( Q = 500\times4.186 = 2093 \) J.
2. Work done on the system, so \( W_{by} = -100 \) J.
3. \( \Delta U = Q - W_{by} = 2093 + 100 = 2193 \) J.

Teacher's Note:
1. Work done on the system adds to its internal energy.
2. Convert calories to joules first.

Answer: (c) 18 J

Explanation:
1. Heat released: \( Q = -20 \) J. Work done by the gas: \( W = -8 \) J.
2. \( \Delta U = Q - W = -20 + 8 = -12 \) J.
3. \( U_f = 30 - 12 = 18 \) J.

Teacher's Note:
1. Heat given out has a negative sign.
2. Work done on the gas has a negative sign in Q = ΔU + W.

Answer: (d) 24 J

Explanation:
1. Along acb: \( \Delta U = Q - W = 200 - 80 = 120 \) J.
2. Internal energy is a state function, so \( \Delta U = 120 \) J along adb as well.
3. \( W = Q - \Delta U = 144 - 120 = 24 \) J.

Teacher's Note:
1. Change in internal energy depends only on the end states.
2. Use it to find the work on the second path.

Answer: (b) −172 J

Explanation:
1. \( \Delta U_{ab} = 200 - 80 = 120 \) J, so \( \Delta U_{ba} = -120 \) J.
2. Work done by the system on path ba is \( W = -52 \) J.
3. \( Q = \Delta U + W = -120 - 52 = -172 \) J, so 172 J of heat is released.

Teacher's Note:
1. Reversing the path reverses the change in internal energy.

Answer: (d) 160 J

Explanation:
1. \( U_b = U_a + \Delta U_{ab} = 40 + 120 = 160 \) J.

Teacher's Note:
1. Internal energy is a state function.

Answer: (d) 13 kJ

Explanation:
1. First path: \( \Delta U = Q - W = 16 - 20 = -4 \) kJ.
2. Second path has the same \( \Delta U \): \( W = Q - \Delta U = 9 + 4 = 13 \) kJ.

Teacher's Note:
1. State functions do not depend on the path.
2. Apply the first law to each path separately.

Answer: (c) 80 J

Explanation:
1. \( \Delta U_{AC} = 90 - 30 = 60 \) J.
2. Along ADC: \( Q = \Delta U + W = 60 + 20 = 80 \) J.

Teacher's Note:
1. Both paths connect the same end states, so ΔU is the same.

Answer: (b) 19.84 cal

Explanation:
1. \( \Delta U = nC_V\Delta T = 2\times4.96\times2 = 19.84 \) cal.

Teacher's Note:
1. For an ideal gas ΔU = nC_VΔT for any process.

Answer: (d) depends on the path

Explanation:
1. Work is the area under the P–V curve, which differs for different paths between the same two states.
2. So work is a path function.

Teacher's Note:
1. Internal energy is a state function; heat and work are path functions.

Answer: (b) 70 J

Explanation:
1. \( W = Q - \Delta U = 110 - 40 = 70 \) J.

Teacher's Note:
1. The remaining heat goes into work.

Answer: (b) \( Q = W + mC_VdT \)

Explanation:
1. The first law states that heat supplied equals work done by the system plus the increase in internal energy.
2. \( Q = W + mC_VdT \).

Teacher's Note:
1. W is work done by the gas in this form of the law.

Answer: (b) \( 6P_0V_0 \)

Explanation:
1. The work equals the area of the triangle.
2. Base \( = 5V_0 - 2V_0 = 3V_0 \), height \( = 5P_0 - P_0 = 4P_0 \).
3. Area \( = \dfrac{1}{2}\times3V_0\times4P_0 = 6P_0V_0 \).
4. The expansion C to A occurs at higher pressure than the compression B to C, so the net work is positive.

Teacher's Note:
1. The loop is traversed so that the expansion is on the upper path, giving positive work.

Answer: (d) 6 J

Explanation:
1. Along iaf: \( \Delta U = 50 - 20 = 30 \) J.
2. The same \( \Delta U \) applies to ibf: \( W = 36 - 30 = 6 \) J.

Teacher's Note:
1. The initial internal energy value of 10 J is not needed for ΔU.

Answer: (c) 4.0 J/K

Explanation:
1. At constant volume \( \Delta U = Q = C_V^{total}\Delta T \).
2. \( C_V^{total} = \dfrac{80}{20} = 4 \) J/K.

Teacher's Note:
1. The question asks for total heat capacity, not molar.

Answer: (d) 1.4×10⁵ N/m²

Explanation:
1. Adiabatic bulk modulus \( = \gamma P \).
2. For a diatomic gas \( \gamma = 1.4 \): \( 1.4\times1.013\times10^5 \approx 1.4\times10^5 \) N/m².

Teacher's Note:
1. Isothermal bulk modulus equals P; adiabatic equals γP.

Answer: (b) \( P \)

Explanation:
1. \( B = -V\dfrac{dP}{dV} \).
2. For isothermal change \( PV = c \), so \( \dfrac{dP}{dV} = -\dfrac{P}{V} \) and \( B = P \).

Teacher's Note:
1. The isothermal bulk modulus is simply the pressure.

Answer: (c) \( -\gamma\,\dfrac{\Delta V}{V} \)

Explanation:
1. \( PV^\gamma = \text{constant} \).
2. Differentiating, \( V^\gamma\,dP + \gamma PV^{\gamma-1}dV = 0 \).
3. \( \dfrac{dP}{P} = -\gamma\dfrac{dV}{V} \).

Teacher's Note:
1. For an isothermal change the factor γ is absent.

Answer: (b) 375 K

Explanation:
1. \( TV^{\gamma-1} = \text{constant} \).
2. \( T_2 = 300\times\left(\dfrac{27}{8}\right)^{2/3} = 300\times\dfrac{9}{4} = 675 \) K.
3. Rise \( = 675 - 300 = 375 \) K.

Teacher's Note:
1. The question asks for the rise, not the final temperature.

Answer: (d) \( \dfrac{5}{2} \)

Explanation:
1. Adiabatic: \( P^{1-\gamma}T^{\gamma} = \text{constant} \), so \( P \propto T^{\gamma/(\gamma-1)} \).
2. For \( \gamma = \dfrac{5}{3} \): \( C = \dfrac{5/3}{2/3} = \dfrac{5}{2} \).

Teacher's Note:
1. Use γ = 5/3 for a monoatomic gas.