THERMODYNAMICS-Exercise

Q1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is \(4.0 × 10^4\ J/g\) ?

Solution

The rate of flow of water through the geyser is given as 3.0 litres per minute. Since the density of water is approximately 1 kg per litre, this corresponds to a mass flow rate of 3 kg per minute.

The initial temperature of water is 27 °C and the final temperature is 77 °C. Hence, the rise in temperature is \(\Delta T = 77 - 27 = 50\,^\circ\text{C}\). The specific heat capacity of water is taken as \(c = 4.2 \times 10^{3}\,\text{J kg}^{-1}\,^\circ\text{C}^{-1}\).

The heat required per minute to raise the temperature of the flowing water is calculated using the relation \(Q = mc\Delta T\).

$$ \begin{aligned} Q &= m c \Delta T \\ &= 3 \times 4.2 \times 10^{3} \times 50 \\ &= 6.3 \times 10^{5}\ \text{J per minute} \end{aligned} $$

The heat of combustion of the fuel used in the gas burner is given as \(4.0 \times 10^{4}\,\text{J g}^{-1}\). Therefore, the mass of fuel consumed per minute is obtained by dividing the heat required per minute by the heat released per gram of fuel.

$$ \begin{aligned} \text{Fuel consumption rate} &= \frac{6.3 \times 10^{5}}{4.0 \times 10^{4}} \\ &= 15.75\ \text{g min}^{-1} \end{aligned} $$

Thus, the rate of consumption of the fuel by the geyser is approximately 15.75 grams per minute.

Q2. What amount of heat must be supplied to \(\mathrm{2.0 × 10^{–2}}\) kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = \(\mathrm{8.3\ J\ mol^{–1}\ K^{–1}}\).)

Solution

The given mass of nitrogen gas is \(2.0 \times 10^{-2}\,\text{kg}\), which is equal to 20 g. The rise in temperature is \(\Delta T = 45\,^\circ\text{C}\). The molecular mass of nitrogen \((\mathrm{N_2})\) is 28 g mol\(^{-1}\), and the universal gas constant is \(R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}\).

The number of moles of nitrogen is obtained by dividing the given mass by its molar mass.

$$ \mu = \frac{20}{28}\ \text{mol} $$

Since nitrogen is a diatomic gas and the process occurs at constant pressure, the molar specific heat capacity at constant pressure is given by \(C_p = \frac{7}{2}R\).

The heat supplied to the gas is calculated using the relation \(Q = \mu C_p \Delta T\).

$$ \begin{aligned} Q &= \mu C_p \Delta T \\ &= \frac{20}{28} \times \frac{7}{2} \times 8.3 \times 45 \\ &= \frac{5}{7} \times \frac{7}{2} \times 8.3 \times 45 \\ &= 933.75\ \text{J} \end{aligned} $$

Thus, the amount of heat that must be supplied to the nitrogen gas at constant pressure is approximately 934 J.

Q3. Explain why
(a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Solution

When two bodies at different temperatures \(T_1\) and \(T_2\) are brought into thermal contact, the final equilibrium temperature depends on the masses and specific heat capacities of the bodies, and not merely on their initial temperatures. The condition for thermal equilibrium is that the heat lost by the hotter body equals the heat gained by the colder body. If the two bodies have unequal heat capacities, the amount of heat exchanged per degree change in temperature will differ, and therefore the equilibrium temperature will shift closer to the temperature of the body with the larger heat capacity. Hence, the final temperature is not necessarily the arithmetic mean \((T_1 + T_2)/2\).

In chemical and nuclear plants, the coolant is required to absorb a large amount of heat without undergoing a significant rise in temperature. A substance with high specific heat can absorb more heat energy per unit mass for a given temperature increase. This property allows the coolant to efficiently remove excess heat from critical components, maintaining safe operating temperatures and preventing thermal damage or overheating of the system.

During driving, the air inside a car tyre undergoes repeated compression and friction due to continuous deformation of the tyre while in contact with the road. This mechanical work increases the internal energy of the air, raising its temperature. Since the tyre is nearly airtight and its volume changes very little, the increase in temperature leads to an increase in pressure, in accordance with the gas laws.

A harbour town experiences a more moderate climate compared to a desert town at the same latitude due to the presence of a large water body nearby. Water has a high specific heat, which enables it to absorb and release large amounts of heat with only small changes in temperature. As a result, the sea heats up slowly during the day and cools down slowly at night, reducing extreme temperature variations. In contrast, desert regions, with low moisture and sandy surfaces of low specific heat, heat up and cool down rapidly, leading to more extreme climatic conditions.

Q4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ?

Solution

The walls of the cylinder are thermally insulating and the piston is also insulated due to the pile of sand placed on it. Hence, no heat is exchanged with the surroundings during compression. The process is therefore adiabatic.

For an adiabatic process, the relation between pressure and volume of a gas is given by

$$ \begin{aligned} P_1 V_1^{\gamma} &= P_2 V_2^{\gamma} \end{aligned} $$

Hydrogen is a diatomic gas, so the ratio of specific heats is

$$ \begin{aligned} \gamma &= \frac{C_p}{C_v} = \frac{7}{5} = 1.4 \end{aligned} $$

Rewriting the adiabatic relation to find the pressure ratio, we obtain

$$ \begin{aligned} \frac{P_2}{P_1} &= \left(\frac{V_1}{V_2}\right)^{\gamma} \end{aligned} $$

The gas is compressed to half of its original volume, so

$$ \begin{aligned} V_2 &= \frac{1}{2} V_1 \end{aligned} $$

Substituting this value into the pressure ratio expression,

$$ \begin{aligned} \frac{P_2}{P_1} &= \left(\frac{V_1}{V_1/2}\right)^{1.4} \\ &= (2)^{1.4} \\ &\approx 2.64 \end{aligned} $$

Thus, when the gas is compressed to half of its original volume under adiabatic conditions, its pressure increases by a factor of approximately 2.64.

Q5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J)

Solution

The work done on the system during the adiabatic change from state A to state B is given as 22.3 J. For an adiabatic process, no heat is exchanged with the surroundings, so the heat absorbed by the system is zero.

$$ \begin{aligned} \Delta U &= Q_1 - W \\ &= 0 - (-22.3) \\ &= +22.3\ \text{J} \end{aligned} $$

Thus, the change in internal energy of the gas between states A and B is 22.3 J. Since internal energy is a state function, this change remains the same for any process connecting the same initial and final states.

In the second process from A to B, the net heat absorbed by the system is given as 9.35 cal. Converting this heat into joules using the relation \(1\,\text{cal} = 4.19\,\text{J}\), we obtain

$$ \begin{aligned} Q_2 &= 9.35 \times 4.19 \\ &= 39.2\ \text{J} \end{aligned} $$

Applying the first law of thermodynamics for the second process,

$$ \begin{aligned} \Delta U &= Q_2 - W \\ 22.3 &= 39.2 - W \\ W &= 39.2 - 22.3 \\ &= 16.9\ \text{J} \end{aligned} $$

Therefore, in the second process, the net work done by the system is 16.9 J.

Q6. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following :
(a) What is the final pressure of the gas in A and B ?
(b) What is the change in internal energy of the gas ?
(c) What is the change in the temperature of the gas ?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ?

Solution

Initially, cylinder A contains a gas at standard temperature and pressure, while cylinder B is completely evacuated. Both cylinders have equal volumes and the entire arrangement is thermally insulated. When the stopcock is suddenly opened, the gas undergoes free expansion into the evacuated cylinder B.

Since the expansion takes place into a vacuum, the gas does no external work. Also, because the system is thermally insulated, no heat is exchanged with the surroundings. Applying the first law of thermodynamics,

$$ \begin{aligned} \Delta U &= Q - W \\ &= 0 - 0 \\ &= 0 \end{aligned} $$

Thus, the change in internal energy of the gas is zero. For an ideal gas, internal energy depends only on temperature. Hence, a zero change in internal energy implies that there is no change in temperature of the gas during the process.

After the stopcock is opened and equilibrium is established, the gas occupies the combined volume of cylinders A and B, which is twice the original volume. Since the temperature remains unchanged, the ideal gas equation gives

$$ \begin{aligned} P_1 V &= P_f (2V) \\ P_f &= \frac{P_1}{2} \end{aligned} $$

Therefore, the final pressure of the gas in both cylinders A and B is half the initial pressure.

During the expansion, the system passes through non-equilibrium states because pressure, temperature, and density are not uniformly defined throughout the gas. Hence, these intermediate states cannot be represented by a single point on the P–V–T surface, which describes only equilibrium states.

Q7. An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

Solution

The electric heater supplies heat energy to the system at a constant rate of 100 W. This means that the rate of heat input to the system is 100 joules per second. The system simultaneously performs work at a rate of 75 joules per second on its surroundings.

According to the first law of thermodynamics, the rate of change of internal energy of a system is equal to the rate of heat supplied to the system minus the rate of work done by the system.

$$ \begin{aligned} \frac{dU}{dt} &= \dot{Q} - \dot{W} \\ &= 100 - 75 \\ &= 25\ \text{W} \end{aligned} $$

Thus, the internal energy of the system is increasing at a rate of 25 joules per second.