GSEB Class 11 Physics Solutions Chapter 12 Thermodynamics

Question 1. 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 fuel if its heat of consumption is 4.0 × 10⁴ J/g?
Answer: Here, the volume of water being heated, \( V = 3.0 \) litre \( \text{min}^{-1} \).
This converts to \( 3 \times 10^{-3} \) \( \text{m}^3 \) \( \text{min}^{-1} \).
The density of water \( \rho = 10^3 \) \( \text{kg} \) \( \text{m}^{-3} \).
So, the mass of water heated, \( m = \rho V \).
\( m = 10^3 \times 3 \times 10^{-3} \) \( \text{kg} \) \( \text{min}^{-1} \).
\( m = 3 \) \( \text{kg} \) \( \text{min}^{-1} \).
The rise in water temperature, \( \Delta\theta = \theta_2 - \theta_1 = 77 - 27 = 50^\circ\text{C} \).
The specific heat capacity of water, \( C = 4.2 \) \( \text{J} \) \( \text{g}^{-1}\text{C}^{-1} \).
This is equivalent to \( 4.2 \times 10^3 \) \( \text{J} \) \( \text{kg}^{-1}\text{C}^{-1} \).
Heat required by water \( = mC\Delta\theta \).
\( = 3 \times 4.2 \times 10^3 \times 50 \) \( \text{J} \) \( \text{min}^{-1} \).
\( = 63 \times 10^4 \) \( \text{J} \) \( \text{min}^{-1} \). (1)
The heat of combustion of the fuel is \( 4.0 \times 10^4 \) \( \text{J} \) \( \text{g}^{-1} \), which is \( 4.0 \times 10^7 \) \( \text{J} \) \( \text{kg}^{-1} \).
Let \( m' \) \( \text{kg} \) \( \text{min}^{-1} \) represent the fuel's combustion rate.
So, the heat supplied by the gas burner from fuel combustion \( = m' \times 4 \times 10^7 \) \( \text{J/min} \). (2)
From equations (1) and (2), we find:
\( 63 \times 10^4 = m' \times 4 \times 10^7 \).
\( m' = \frac{63 \times 10^4}{4 \times 10^7} \).
\( = 15.75 \times 10^{-3} \) \( \text{kg} \) \( \text{min}^{-1} \).
Thus, the fuel's rate of combustion is \( 16 \) \( \text{g} \) \( \text{min}^{-1} \).
In simple words: First, we calculated the total heat needed to warm the water. Then, we used the fuel's energy value to figure out how much fuel would be burned to provide that much heat.

Exam Tip: Remember to convert all units to a consistent system (like SI units) before performing calculations. Pay close attention to unit conversions for mass (grams to kilograms) and specific heat capacity.

Question 2. What amount of heat must be supplied to 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 = 8.3 J mol^-1K^-1).
Answer: Here, the mass of the gas, \( m = 2.0 \times 10^{-2} \) \( \text{kg} = 20 \) \( \text{g} \).
The rise in temperature, \( \Delta T = 45^\circ\text{C} \).
The molar mass of \( \text{N}_2 \), \( M = 28 \) \( \text{g/mol} \).
If \( n \) is the number of moles in \( m \), then:
\( n = \frac{\text{Mass in gram}}{\text{Molecular mass}} = \frac{m}{M} \).
\( n = \frac{20}{28} = 0.714 \).
The gas constant \( R = 8.3 \) \( \text{J} \) \( \text{mol}^{-1}\text{K}^{-1} \).
If \( C_p \) is the molar specific heat of the gas at constant pressure, then for a diatomic gas like nitrogen:
\( C_p = \frac{7}{2}R = \frac{7}{2} \times 8.3 \) \( \text{J} \) \( \text{mol}^{-1}\text{K}^{-1} \).
We need to find the heat supplied, \( Q \).
Using the relation \( Q = nC_p\Delta\theta \), we get:
\( Q = nC_p\Delta T \).
\( = 0.714 \times \frac{7}{2} \times 8.3 \times 45 \) \( \text{J} \).
\( = 933.75 \) \( \text{J} = 934 \) \( \text{J} \).
In simple words: We calculated the number of moles of nitrogen, then used the specific heat at constant pressure to find the total heat needed to increase its temperature by the given amount.

Exam Tip: Always identify whether the process occurs at constant volume or constant pressure, as this determines whether to use \( C_v \) or \( C_p \) for specific heat calculations. Remember to convert mass to moles.

Question 3. Explain the following statements regarding thermal physics:
(a) Two bodies at different temperatures \( T_1 \) and \( T_2 \) if brought in thermal contact don't necessarily settle to the mean temperature \( (T_1 + T_2)/2 \).
(b) The coolant in a chemical or nuclear plant (i.e, 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.
Answer:
(a) This happens because the two bodies might have different masses and also different specific heat capacities. When in thermal contact, heat moves from the hotter body to the colder body until their temperatures become equal. The final temperature can only be the average temperature, i.e., \( \frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2} \), if the thermal capacities of both bodies are identical.
(b) The function of the coolant in a chemical or nuclear plant is to take in the heat produced and thus stop its various parts from becoming too hot. It possesses a high specific heat capacity. The heat that a substance absorbs is directly proportional to its specific heat. So, a higher specific heat value means more heat will be absorbed by the coolant for the same temperature rise, protecting the plant from overheating.
(c) The temperature of the tyre and the air inside it rises during driving because of friction between the tyres and the road. The volume of the tyre generally stays unchanged. Based on Charles's law (which relates pressure and temperature: \( P \propto T \)), the air pressure inside the tyre goes up as the temperature increases.
(d) This is due to the presence of a large amount of water vapour in the air, which results from the town's closeness to the sea.
In simple words: (a) Objects don't always reach an average temperature when touching because they might be different sizes or made of different materials. (b) Coolants need to absorb a lot of heat without getting too hot themselves, so they need high specific heat. (c) Car tires get warmer while driving, which makes the air inside them push harder, increasing pressure. (d) Coastal towns have milder weather because the sea adds moisture to the air.

Exam Tip: For conceptual questions, provide a clear, concise explanation, stating the underlying principle or law where applicable. Use examples to illustrate your point if helpful.

Question 4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made Of 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?
Answer: Here, let the initial volume be \( V_1 = X \).
The final volume is \( V_2 = \frac{V_1}{2} = \frac{X}{2} \).
For a diatomic gas like hydrogen, the ratio of specific heats \( \gamma = \frac{C_p}{C_v} \).
\( \gamma = \frac{7}{5} = 1.4 \).
We need to find the factor by which pressure increases, which is \( \frac{P_2}{P_1} = ? \).
Since no heat can be exchanged (insulated walls and piston), the process is adiabatic.
For an adiabatic change, we use the relation:
\( P_1V_1^\gamma = P_2V_2^\gamma \).
This means \( \frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^\gamma \).
Substituting the values:
\( \frac{P_2}{P_1} = \left(\frac{X}{X/2}\right)^{1.4} \).
\( = (2)^{1.4} = 2.64 \).
In simple words: Because the gas is compressed without heat loss, its pressure will rise by a certain amount. We used the adiabatic gas law to calculate that the pressure increases by a factor of 2.64 when the volume is halved.

Exam Tip: Identify the type of thermodynamic process (isothermal, adiabatic, isobaric, isochoric) correctly, as this determines which formula to use. Remember the \( \gamma \) value for different types of gases.

Question 5. 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)?
Answer: Here, when the gas changes its state adiabatically from A to B, the work done on the system is used to increase the gas's internal energy.
So, the work done, \( dW = 22.3 \) \( \text{J} \).
If \( dU \) represents the increase in internal energy, then:
\( dU = dW = 22.3 \) \( \text{J} \). (i)
In the second case, when the gas moves from state A to state B, the heat absorbed by the system \( \Delta Q \) is:
\( \Delta Q = 9.35 \) \( \text{cal} \).
Converting to Joules: \( \Delta Q = 9.35 \times 4.19 \) \( \text{J} \) (since \( 1 \) \( \text{cal} = 4.19 \) \( \text{J} \)).
\( = 39.18 \) \( \text{J} \).
Let \( \Delta W' \) be the net work done by the system. Then, using the first law of thermodynamics:
\( \Delta Q = \Delta U + \Delta W \).
We rearrange this to find the work done:
\( \Delta W' = \Delta Q - \Delta U \).
\( = 39.18 - 22.3 \).
\( = 16.88 \) \( \text{J} \).
\( = 16.9 \) \( \text{J} \).
In simple words: We first found the change in internal energy from the adiabatic process. Then, using the first law of thermodynamics, we subtracted this internal energy change from the heat absorbed in the second process to find the new work done.

Exam Tip: The first law of thermodynamics, \( \Delta Q = \Delta U + \Delta W \), is crucial. Pay attention to the signs: work done *on* the system is negative \( \Delta W \), work done *by* the system is positive \( \Delta W \). Heat absorbed is positive \( \Delta Q \).

Question 6. 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 finally 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 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?'
Answer:
(a) As the initial and final temperatures remain the same, when the stopcock is suddenly opened, the volume available to the gas at 1 atmosphere pressure will become twice its initial value. Therefore, the pressure will decrease to half, which means 0.5 atmosphere.
Using \( P_2V_2 = P_1V_1 \).
Here, initial pressure \( P_1 = 1 \) \( \text{atm} \).
Initial volume \( V_1 = V \).
Final volume \( V_2 = 2V \).
We need to find final pressure \( P_2 \).
\( P_2 = \frac{P_1 V_1}{V_2} \).
\( = \frac{1 \times V}{2V} \).
\( = 0.5 \) \( \text{atm} \).
(b) The change in internal energy is zero. This is because no work is done on or by the gas, so there will be no change in its internal energy under constant temperature conditions.
(c) If we assume the gas is ideal, then there is no change in its temperature, because the gas performs no work during its expansion, i.e., \( \Delta T = 0 \).
(d) No, because this process (known as free expansion) is very quick and uncontrolled. The intermediate states of the system (the gas in this case) are non-equilibrium states and do not satisfy the gas equation. So, they do not lie on the P-V-T surface. Eventually, the gas will return to an equilibrium state that does lie on its P-V-T surface.
In simple words: (a) When the gas expands into the empty cylinder, its volume doubles, so its pressure halves. (b) No work is done, so the internal energy of the gas doesn't change. (c) For an ideal gas, no work means no temperature change. (d) The quick expansion means the gas isn't in a stable state until it settles, so its properties don't follow the usual gas laws during the expansion itself.

Exam Tip: For ideal gas expansions into a vacuum (free expansion), remember that no work is done and no heat is exchanged, leading to no change in internal energy and temperature. Clearly distinguish between equilibrium and non-equilibrium states.

Question 7. A steam engine delivers 5.4 × 10⁹ J of work per minute and services 3.6 × 10⁹ J of heat per minute from the boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer: Here, the heat absorbed from the boiler per minute, \( Q_1 = 3.6 \times 10^9 \) \( \text{J} \).
The work done by the steam engine per minute, \( W = 5.4 \times 10^8 \) \( \text{J} \).
We need to find the heat wasted or rejected per minute, \( Q_2 \), and the percentage efficiency of the engine, \( \eta \% \).
We know that the efficiency is given by:
\( \eta \% = \frac{W}{Q_1} \times 100 \).
Substituting the values:
\( \eta \% = \frac{5.4 \times 10^8 \text{ J}}{3.6 \times 10^9 \text{ J}} \times 100 \).
\( = \frac{3}{20} \times 100 = 15\% \).
Also, using the relation for heat engines, \( Q_1 = W + Q_2 \), we get:
\( Q_2 = Q_1 - W \).
\( = (3.6 \times 10^9) - (5.4 \times 10^8) \).
\( = (36 \times 10^8) - (5.4 \times 10^8) \).
\( = 30.6 \times 10^8 \) \( \text{J/min} \).
\( = 3.06 \times 10^9 \) \( \text{J/min} \).
Rounding off, this is \( 3.1 \times 10^9 \) \( \text{J/min} \).
In simple words: We calculated how efficiently the engine converts the heat it takes in into useful work. Then, by subtracting the useful work from the total heat input, we found out how much heat energy was simply lost or wasted.

Exam Tip: Remember the efficiency formula for a heat engine (\( \eta = W/Q_H \)) and the relationship between heat input, work, and heat rejected (\( Q_H = W + Q_C \)). Ensure consistent units for all values.

Question 8. An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Answer: Here, the rate of heat supplied per second, \( dQ = 100 \) \( \text{J} \).
The rate of work done by the system per second, \( dW = 75 \) \( \text{J} \).
We need to find the increase in internal energy per second, \( dU \).
According to the first law of thermodynamics, we understand that:
\( dQ = dU + dW \).
Rearranging this to find \( dU \):
\( dU = dQ - dW \).
\( = 100 \) \( \text{J} - 75 \) \( \text{J} \).
\( = 25 \) \( \text{J per second} \).
This is also \( 25 \) \( \text{W} \) (since \( 1 \) \( \text{watt} = 1 \) \( \text{Joule per second} \)).
In simple words: We used the first law of thermodynamics, which states that the energy added as heat equals the work done plus the change in internal energy. By subtracting the work done from the heat supplied, we found how quickly the system's inner energy was growing.

Exam Tip: The first law of thermodynamics (\( \Delta U = Q - W \)) is fundamental. For rates, it becomes \( \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} \). Always check the direction of heat flow and work done to apply correct signs.

Question 9. A thermodynamic system is taken from an original state to an intermediate state by linear process shown in figure here. Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F?
Answer: Let \( W \) be the total work done by the gas when moving from D to E and then to F.
If \( W_1 \) and \( W_2 \) represent the work done by the gas from D to E and E to F, respectively, then:
\( W = W_1 + W_2 \). (1)
Now, \( W_1 \) is the work done during the process from D to E (expansion), which equals the area under the curve DEHGD.
This area can be calculated as the area of rectangle EHGF plus the area of triangle ADEF.
\( W_1 = \text{Area of rectangle EHGD} = \frac{1}{2} \times \text{EF} \times \text{DF} + \text{GH} \times \text{FG} \). (2)
From the figure:
\( \text{EF} = 5 - 2 = 3 \) \( \text{m}^3 \).
\( \text{DF} = 600 - 300 = 300 \) \( \text{Nm}^{-2} \). (3)
\( \text{FG} = 300 - 0 = 300 \) \( \text{Nm}^{-2} \).
\( \text{GH} = 5 - 2 = 3 \) \( \text{m}^3 \).
From (2) and (3), we get:
\( W_1 = \left[\frac{1}{2} \times 3 \times 300 + 3 \times 300\right] \) \( \text{J} \). (4)
\( W_1 = [450 + 900] \) J = 1350 J.
\( W_2 \) is the work done during the process from E to F (compression).
This equals the area EHGF.
\( W_2 = - \text{FG} \times \text{GH} \).
\( = - (300 - 0) \times (5 - 2) \).
\( = - 300 \times 3 \) \( \text{J} \). (5)
\( = - 900 \) \( \text{J} \).
From (1), (4), and (5), we get:
\( W = 1350 + (-900) \).
\( W = 450 \) \( \text{J} \).
In simple words: We broke the process into two parts. First, we calculated the work done during expansion by finding the area under that part of the graph. Then, we calculated the work done during compression. The total work is the sum of these two amounts.

Exam Tip: For P-V diagrams, work done is the area under the curve. Remember that work done *by* the gas during expansion is positive, and work done *on* the gas during compression is negative.

Question 10. A refrigerator is to remove heat from the eatable kept inside at 9°C. Calculate the coefficient of performance if the room temperature is 36°C?
Answer: Here, the room temperature (hot reservoir) \( T_1 = 273 + 36 \).
\( T_1 = 309 \) \( \text{K} \).
The temperature inside the refrigerator (cold reservoir) \( T_2 = 9^\circ\text{C} \).
\( T_2 = 282 \) \( \text{K} \).
We need to calculate the coefficient of performance, \( \beta \).
Using the relation for the coefficient of performance of a refrigerator:
\( \beta = \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}-\mathrm{T}_{2}} \).
Substituting the temperature values, we get:
\( \beta = \frac{282}{309-282} = \frac{282}{27} \).
\( \beta = 10.4 \).
In simple words: We converted the given temperatures to Kelvin. Then, we used the formula for a refrigerator's coefficient of performance, which compares the cold temperature to the temperature difference, to find the answer.

Exam Tip: Always convert temperatures to Kelvin when using thermodynamic formulas like the coefficient of performance. Understand that a higher coefficient of performance indicates a more efficient refrigerator.