ThermodynamicsClass 11 Physics NCERT Solutions

Calculation: First, calculate the mass of water flowing per minute: $m = 3.0 \frac{\text{litres}}{\text{min}} \times 1000 \frac{\text{g}}{\text{litre}} = 3000 \frac{\text{g}}{\text{min}}$

Next, calculate the change in temperature: $\Delta T = T_2 - T_1 = 77^{\circ} \mathrm{C} - 27^{\circ} \mathrm{C} = 50^{\circ} \mathrm{C} = 50 \text{ K}$

Now, calculate the heat required per minute to heat this water: $\Delta Q = m s \Delta T = (3000 \text{ g}) \times (4.2 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}) \times (50 \text{ K})$ $\Delta Q = 630000 \text{ J/min} = 6.3 \times 10^5 \text{ J/min}$

Finally, calculate the rate of fuel consumption. This is the mass of fuel required to produce this amount of heat per minute. $\text{Rate of consumption} = \frac{\Delta Q}{\text{Heat of combustion}} = \frac{6.3 \times 10^5 \text{ J/min}}{4.0 \times 10^4 \text{ J/g}}$ $\text{Rate of consumption} = 15.75 \text{ g/min}$

Final Answer: The rate of consumption of the fuel is $15.75 \text{ g/min}$.

Calculation: First, calculate the number of moles of nitrogen: $n = \frac{m}{M} = \frac{20 \text{ g}}{28 \text{ g/mol}} = \frac{5}{7} \text{ mol} \approx 0.714 \text{ mol}$

Next, calculate the molar specific heat capacity at constant pressure for nitrogen (a diatomic gas): $C_p = \frac{7}{2} R = \frac{7}{2} \times 8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} = 29.05 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Now, calculate the amount of heat supplied: $\Delta Q = n C_p \Delta T = \left(\frac{5}{7} \text{ mol}\right) \times (29.05 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) \times (45 \text{ K})$ $\Delta Q = 933.4 \text{ J}$

Final Answer: The amount of heat that must be supplied is $933.4 \text{ J}$.

(a) When two bodies at different temperatures $T_1$ and $T_2$ are brought into thermal contact, heat flows from the hotter body to the colder body until they reach a common equilibrium temperature, $T_f$. According to the principle of calorimetry, the heat lost by the hot body is equal to the heat gained by the cold body. Let the masses be $m_1$ and $m_2$, and specific heat capacities be $s_1$ and $s_2$. Assuming $T_1 > T_2$, we have: $m_1 s_1 (T_1 - T_f) = m_2 s_2 (T_f - T_2)$ Solving for $T_f$ gives: $T_f = \frac{m_1 s_1 T_1 + m_2 s_2 T_2}{m_1 s_1 + m_2 s_2}$ The final temperature $T_f$ will be the mean temperature $(T_1 + T_2)/2$ only in the special case where their heat capacities are equal, i.e., $m_1 s_1 = m_2 s_2$. In general, the masses and specific heat capacities are different, so the final temperature will not be the simple arithmetic mean.

(b) A coolant is used to absorb excess heat generated in a plant to prevent overheating. The amount of heat $\Delta Q$ a substance can absorb for a given temperature rise $\Delta T$ is given by $\Delta Q = m s \Delta T$. This can be rewritten as $\Delta T = \frac{\Delta Q}{ms}$. For a given mass of coolant ($m$) and a certain amount of heat to be removed ($\Delta Q$), the rise in its temperature ($\Delta T$) is inversely proportional to its specific heat capacity ($s$). A coolant with a high specific heat capacity will show a smaller temperature increase while absorbing a large amount of heat. This makes it more effective at transferring heat away from the hot parts of the plant without boiling or undergoing large temperature changes itself.

(c) During driving, a car tyre constantly flexes and deforms, and there is friction between the tyre and the road surface. This work done against frictional forces and internal friction within the tyre's material is converted into heat. This heat increases the internal energy and thus the temperature of the air inside the tyre. According to the ideal gas law, $PV = nRT$. Since the volume of the tyre ($V$) is nearly constant, the pressure ($P$) of the air inside is directly proportional to its absolute temperature ($T$). As the temperature of the air increases, its pressure also increases.

(d) The climate of a harbour town is more temperate (meaning less extreme temperature variations between day and night, and summer and winter) because of its proximity to a large body of water (the sea). Water has a very high specific heat capacity (approximately $4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$). This means it takes a large amount of heat to raise its temperature and it releases a large amount of heat when it cools. During the day, the water absorbs a large amount of solar energy without a significant rise in temperature, keeping the coastal area cool. At night, the water cools down slowly, releasing this stored heat and keeping the area warmer than inland regions. In contrast, a desert town is surrounded by land (sand and rock), which has a low specific heat capacity. The land heats up very quickly during the day and cools down very quickly at night, leading to extreme temperature differences.

Formula: For an adiabatic process, the relation between pressure and volume is given by: $P V^{\gamma} = \text{constant}$ $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$ where $\gamma = \frac{C_p}{C_v}$ is the ratio of specific heats.

Calculation: Hydrogen ($H_2$) is a diatomic gas. For a diatomic gas, the molar specific heat at constant volume is $C_v = \frac{5}{2}R$ and at constant pressure is $C_p = \frac{7}{2}R$. Therefore, the ratio of specific heats is: $\gamma = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} = 1.4$

Using the adiabatic relation: $P_1 V_1^{1.4} = P_2 V_2^{1.4}$ We need to find the ratio $\frac{P_2}{P_1}$: $\frac{P_2}{P_1} = \frac{V_1^{1.4}}{V_2^{1.4}} = \left(\frac{V_1}{V_2}\right)^{1.4}$ Given that $V_2 = \frac{1}{2} V_1$, so $\frac{V_1}{V_2} = 2$. $\frac{P_2}{P_1} = (2)^{1.4}$ To calculate $2^{1.4}$, we can use logarithms or a calculator: $\log\left(\frac{P_2}{P_1}\right) = 1.4 \log(2) = 1.4 \times 0.3010 = 0.4214$ $\frac{P_2}{P_1} = \text{antilog}(0.4214) \approx 2.639$

Final Answer: The pressure of the gas increases by a factor of approximately $2.64$.

Calculation: First, we determine the change in internal energy ($\Delta U$) when the system goes from state A to state B. Internal energy is a state function, meaning its change depends only on the initial and final states (A and B), not on the path taken. We can calculate $\Delta U$ using the information from the first (adiabatic) process.

For the adiabatic process from A to B: $\Delta U = \Delta Q_1 - \Delta W_1$ $\Delta U = 0 - (-22.3 \text{ J}) = 22.3 \text{ J}$ So, the change in internal energy in going from state A to B is $22.3 \text{ J}$.

Now, we consider the second process, where the gas is taken from A to B via a different path. The change in internal energy $\Delta U$ will be the same. $\Delta U = 22.3 \text{ J}$ Heat absorbed in this process is $\Delta Q_2 = 9.35 \text{ cal}$. Let's convert this to Joules: $\Delta Q_2 = 9.35 \text{ cal} \times 4.19 \text{ J/cal} = 39.1765 \text{ J}$

Now, we apply the First Law of Thermodynamics to this second process to find the work done by the system, $\Delta W_2$: $\Delta Q_2 = \Delta U + \Delta W_2$ $\Delta W_2 = \Delta Q_2 - \Delta U$ $\Delta W_2 = 39.1765 \text{ J} - 22.3 \text{ J} = 16.8765 \text{ J}$

Final Answer: The net work done by the system in the latter case is approximately $16.9 \text{ J}$.

This process describes the free expansion of a gas.

When the stopcock is opened, the gas expands to fill both cylinders. The final volume is $V_2 = V + V = 2V$.

The expansion is into a vacuum, so the gas does no external work. $\Delta W = 0$.

(a) What is the final pressure of the gas in A and B? From the First Law of Thermodynamics, $\Delta U = \Delta Q - \Delta W$. Since $\Delta Q = 0$ and $\Delta W = 0$, the change in internal energy is $\Delta U = 0$. For an ideal gas, internal energy depends only on temperature. Since $\Delta U = 0$, the temperature of the gas does not change, i.e., $T_2 = T_1$. We can now use Boyle's Law ($P_1 V_1 = P_2 V_2$) since the temperature is constant. $P_1 V_1 = P_2 V_2$ $(1 \text{ atm}) (V) = P_2 (2V)$ $P_2 = \frac{1 \text{ atm} \times V}{2V} = 0.5 \text{ atm}$ The final pressure of the gas is $0.5$ atm.

(b) What is the change in internal energy of the gas? As explained above, the system is thermally insulated ($\Delta Q=0$) and the gas expands into a vacuum, so no work is done ($\Delta W=0$). From the First Law of Thermodynamics, $\Delta U = \Delta Q - \Delta W = 0 - 0 = 0$. The change in internal energy of the gas is zero.

(c) What is the change in the temperature of the gas? For an ideal gas, the internal energy is a function of temperature only ($U \propto T$). Since the change in internal energy $\Delta U = 0$, the change in temperature $\Delta T$ must also be zero. The final temperature is the same as the initial temperature.

(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface? No. The P-V-T surface represents a collection of equilibrium states. The free expansion of the gas is a rapid and irreversible process. During the expansion, the gas is not in thermodynamic equilibrium. Variables like pressure and temperature are not well-defined for the system as a whole; they may vary from point to point within the gas. Since the intermediate states are non-equilibrium states, they cannot be represented by points on the P-V-T surface of the system.

Formula: The First Law of Thermodynamics is given by $\Delta Q = \Delta U + \Delta W$. To find the rate of change, we can differentiate this equation with respect to time: $\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$

Calculation: We rearrange the formula to solve for the rate of change of internal energy: $\frac{\Delta U}{\Delta t} = \frac{\Delta Q}{\Delta t} - \frac{\Delta W}{\Delta t}$ Substituting the given values: $\frac{\Delta U}{\Delta t} = 100 \text{ J/s} - 75 \text{ J/s}$ $\frac{\Delta U}{\Delta t} = 25 \text{ J/s} = 25 \text{ W}$

Final Answer: The internal energy is increasing at a rate of $25 \text{ J/s}$ (or 25 Watts).

Calculation: First, calculate the work done during the process D $\rightarrow$ E (expansion): $W_{DE} = \frac{1}{2} (600 \text{ N/m}^2 + 300 \text{ N/m}^2) (5.0 \text{ m}^3 - 2.0 \text{ m}^3)$ $W_{DE} = \frac{1}{2} (900 \text{ N/m}^2) (3.0 \text{ m}^3)$ $W_{DE} = 450 \times 3.0 \text{ J} = 1350 \text{ J}$ Since the volume increases, the work done by the gas is positive.

Next, calculate the work done during the isobaric process E $\rightarrow$ F (compression): $W_{EF} = P_E (V_F - V_E) = 300 \text{ N/m}^2 (2.0 \text{ m}^3 - 5.0 \text{ m}^3)$ $W_{EF} = 300 \text{ N/m}^2 (-3.0 \text{ m}^3)$ $W_{EF} = -900 \text{ J}$ Since the volume decreases, the work done by the gas is negative (work is done on the gas).

Finally, calculate the total work done by the gas: $W_{DEF} = W_{DE} + W_{EF}$ $W_{DEF} = 1350 \text{ J} + (-900 \text{ J})$ $W_{DEF} = 450 \text{ J}$

Final Answer: The total work done by the gas from D to E to F is $450 \text{ J}$.