Thermodynamics — NCERT Solutions

Given:

• Work done on the system in adiabatic process (A → B): $W_{on} = 22.3$ J
• So work done by the system adiabatically: $W_{adiabatic} = -22.3$ J
• Heat absorbed in the second process (A → B): $\Delta Q = 9.35$ cal
• $1$ cal $= 4.19$ J

Step 1: Find change in internal energy $\Delta U$ using the adiabatic process

For an adiabatic process, $\Delta Q = 0$. By the First Law:
$$\Delta Q = \Delta U + W_{by\ system}$$
$$0 = \Delta U + (-22.3)$$
$$\Delta U = +22.3 \text{ J}$$

Note: Internal energy is a state function, so $\Delta U$ from A to B is the same regardless of the path.

Step 2: Convert heat absorbed in second process to Joules
$$\Delta Q = 9.35 \times 4.19 = 39.18 \text{ J}$$

Step 3: Apply First Law to the second process (A → B)
$$\Delta Q = \Delta U + W$$
$$39.18 = 22.3 + W$$
$$W = 39.18 - 22.3 = 16.88 \text{ J}$$

Answer: The net work done by the system in the second process is approximately 16.9 J.

Given:

• Cylinder A: gas at STP → $P_1 = 1$ atm, $T_1 = 273$ K, volume $= V$
• Cylinder B: completely evacuated, volume $= V$
• System is thermally insulated (adiabatic walls)
• Stopcock is suddenly opened → free expansion

(a) Final pressure of the gas in A and B:

When the stopcock is opened, the gas expands freely to fill both cylinders. The total volume doubles: $V_{final} = 2V$.

Since the system is thermally insulated and the gas expands against zero external pressure (vacuum in B), no work is done:
$$W = 0, \quad \Delta Q = 0$$

For an ideal gas in free expansion, temperature remains unchanged (shown in part c). Applying Boyle's Law:
$$P_1 V = P_2 (2V)$$
$$P_2 = \frac{P_1}{2} = \frac{1}{2} \text{ atm} = 0.5 \text{ atm}$$

The final pressure in A and B is 0.5 atm.

(b) Change in internal energy of the gas:

By the First Law of Thermodynamics:
$$\Delta U = \Delta Q - W$$

• $\Delta Q = 0$ (thermally insulated system)
• $W = 0$ (gas expands into vacuum, no external pressure to work against)

$$\Delta U = 0 - 0 = 0$$

There is no change in the internal energy of the gas.

(c) Change in temperature of the gas:

For an ideal gas, internal energy depends only on temperature:
$$\Delta U = \mu C_v \Delta T$$

Since $\Delta U = 0$:
$$\mu C_v \Delta T = 0 \implies \Delta T = 0$$

There is no change in the temperature of the gas.

(d) Do the intermediate states lie on the P-V-T surface?

No. The process of free expansion is a sudden, irreversible process. During the intermediate stages, the gas is not in thermodynamic equilibrium — the pressure and temperature are not uniform throughout the gas. Therefore, the intermediate states cannot be described by well-defined values of P, V, and T, and hence they do not lie on the equilibrium P-V-T surface of the system. Only the initial and final equilibrium states lie on the P-V-T surface.

Given:

• Rate of heat supplied to the system: $\dfrac{\Delta Q}{\Delta t} = 100$ W $= 100$ J/s
• Rate of work done by the system: $\dfrac{\Delta W}{\Delta t} = 75$ J/s

Concept: First Law of Thermodynamics:
$$\Delta Q = \Delta U + \Delta W$$

Dividing throughout by $\Delta t$:
$$\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$$

Calculation:
$$\frac{\Delta U}{\Delta t} = \frac{\Delta Q}{\Delta t} - \frac{\Delta W}{\Delta t}$$
$$\frac{\Delta U}{\Delta t} = 100 - 75 = 25 \text{ J/s} = 25 \text{ W}$$

Answer: The internal energy of the system is increasing at the rate of 25 J/s (25 W).

Given (reading from the P-V diagram, Fig. 11.11):

From the figure description, the process D → E is a linear (straight-line) process on a P-V diagram, and E → F is an isobaric (constant pressure) process that returns the volume to its original value.

Reading standard values from the figure:

• At point D: $P_D = 300$ Pa, $V_D = 2$ m³ (initial state)
• At point E: $P_E = 500$ Pa, $V_E = 5$ m³ (intermediate state)
• At point F: $P_F = 500$ Pa, $V_F = 2$ m³ (final state, isobaric from E)

(Note: These are the standard values given in the NCERT figure for this problem.)

Step 1: Work done in process D → E (linear process on P-V diagram)

The process is a straight line on the P-V diagram. Work done = area under the P-V curve.

This area is a trapezium with:

• Parallel sides = $P_D = 300$ Pa and $P_E = 500$ Pa
• Width = $V_E - V_D = 5 - 2 = 3$ m³

$$W_{DE} = \frac{1}{2}(P_D + P_E)(V_E - V_D)$$
$$W_{DE} = \frac{1}{2}(300 + 500)(3)$$
$$W_{DE} = \frac{1}{2} \times 800 \times 3 = 1200 \text{ J}$$

Step 2: Work done in process E → F (isobaric process)

Pressure is constant at $P_E = 500$ Pa. Volume decreases from $V_E = 5$ m³ to $V_F = 2$ m³.

$$W_{EF} = P_E(V_F - V_E) = 500 \times (2 - 5) = 500 \times (-3) = -1500 \text{ J}$$

(Negative because work is done on the gas as volume decreases.)

Step 3: Total work done from D to E to F
$$W_{total} = W_{DE} + W_{EF} = 1200 + (-1500) = -300 \text{ J}$$

Answer: The total work done by the gas from D to E to F is $\mathbf{-300 \text{ J}}$. The negative sign indicates that net work is done on the gas.