Thermodynamics

Given:
- Flow rate of water = 3.0 litres/min = 3.0 kg/min (since density of water = 1 kg/litre)
- Initial temperature, $T_1 = 27°C$
- Final temperature, $T_2 = 77°C$
- Rise in temperature, $\Delta T = 77 - 27 = 50°C$
- Heat of combustion of fuel, $H = 4.0 \times 10^4$ J/g
- Specific heat of water, $s = 4.2 \times 10^3$ J kg⁻¹ K⁻¹

Concept: Heat required = $ms\Delta T$

Step 1: Calculate heat required per minute
$$\Delta Q = ms\Delta T = 3.0 \times 4.2 \times 10^3 \times 50$$
$$\Delta Q = 6.3 \times 10^5 \text{ J/min}$$

Step 2: Calculate rate of fuel consumption

Let the rate of fuel consumption be $r$ g/min.

Heat supplied by fuel per minute = $r \times H = r \times 4.0 \times 10^4$ J/min

Setting heat supplied equal to heat required:
$$r \times 4.0 \times 10^4 = 6.3 \times 10^5$$
$$r = \frac{6.3 \times 10^5}{4.0 \times 10^4} = 15.75 \text{ g/min}$$

Answer: The rate of consumption of fuel is approximately 15.75 g/min.

Given:
- Mass of nitrogen, $m = 2.0 \times 10^{-2}$ kg $= 20$ g
- Rise in temperature, $\Delta T = 45°C = 45$ K
- Molecular mass of $N_2$, $M = 28$ g/mol
- $R = 8.3$ J mol⁻¹ K⁻¹
- Process: constant pressure

Step 1: Find number of moles
$$\mu = \frac{m}{M} = \frac{20}{28} = \frac{5}{7} \text{ mol}$$

Step 2: Find $C_p$ for nitrogen

Nitrogen ($N_2$) is a diatomic gas. For a diatomic gas:
$$C_v = \frac{5}{2}R, \quad C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$$
$$C_p = \frac{7}{2} \times 8.3 = 29.05 \text{ J mol}^{-1}\text{K}^{-1}$$

Step 3: Calculate heat supplied at constant pressure
$$\Delta Q = \mu C_p \Delta T$$
$$\Delta Q = \frac{5}{7} \times 29.05 \times 45$$
$$\Delta Q = \frac{5}{7} \times 1307.25$$
$$\Delta Q = \frac{6536.25}{7} \approx 933.75 \text{ J}$$

Answer: The amount of heat that must be supplied is approximately 933.75 J ≈ 9.34 × 10² J.

(a) Two bodies at different temperatures T₁ and T₂ do not necessarily settle to the mean temperature (T₁ + T₂)/2:

The final equilibrium temperature depends on the masses and specific heat capacities of the two bodies, not just their temperatures. When two bodies exchange heat, the heat lost by the hotter body equals the heat gained by the cooler body:
$$m_1 s_1 (T_1 - T_f) = m_2 s_2 (T_f - T_2)$$
Solving: $T_f = \dfrac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2}$

This equals $\dfrac{T_1+T_2}{2}$ only when $m_1 s_1 = m_2 s_2$. In general, this condition is not satisfied, so the final temperature is not the mean temperature.

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(b) The coolant should have high specific heat:

Specific heat capacity $s$ is defined as the heat absorbed per unit mass per unit rise in temperature:
$$\Delta Q = ms\Delta T$$
A coolant with high specific heat can absorb a large amount of heat from the plant for a small rise in its own temperature. This makes it very effective in carrying away heat and keeping the plant cool without itself becoming too hot. Hence, a high specific heat coolant is preferred.

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(c) Air pressure in a car tyre increases during driving:

During driving, the tyre undergoes repeated deformation due to friction with the road. This friction generates heat, raising the temperature of the air inside the tyre. According to Gay-Lussac's Law (at constant volume):
$$\frac{P}{T} = \text{constant} \implies P \propto T$$
As temperature $T$ increases, the pressure $P$ of the air inside the tyre increases.

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(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude:

Water has a very high specific heat capacity ($\approx 4200$ J kg⁻¹K⁻¹) compared to sand/soil ($\approx 840$ J kg⁻¹K⁻¹). The sea near a harbour town absorbs a large amount of heat in summer without a large rise in temperature, and releases heat slowly in winter. This moderates the temperature of the surrounding coastal area. In contrast, the sand in a desert heats up and cools down rapidly (low specific heat), causing extreme temperature variations. Hence, harbour towns have a more temperate (moderate) climate.

Given:
- Number of moles of hydrogen, $\mu = 3$ mol
- Initial conditions: Standard Temperature and Pressure (STP)
- Walls are heat-insulating and piston is insulated → adiabatic process
- Final volume, $V_2 = \dfrac{V_1}{2}$

Concept: For an adiabatic process:
$$PV^\gamma = \text{constant}$$
$$P_1 V_1^\gamma = P_2 V_2^\gamma$$

Step 1: Find $\gamma$ for hydrogen

Hydrogen ($H_2$) is a diatomic gas:
$$\gamma = \frac{C_p}{C_v} = \frac{7/2\, R}{5/2\, R} = \frac{7}{5} = 1.4$$

Step 2: Calculate the pressure ratio
$$\frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma = \left(\frac{V_1}{V_1/2}\right)^{1.4} = (2)^{1.4}$$

Step 3: Evaluate $(2)^{1.4}$
$$(2)^{1.4} = 2^{7/5} = (2^7)^{1/5} = (128)^{0.2} \approx 2.639$$

Answer: The pressure of the gas increases by a factor of approximately $\mathbf{2^{1.4} \approx 2.64}$.

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

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(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.

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(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.

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(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.

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(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.)*

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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

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

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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.)

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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.