**12.1. 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 the fuel if its heat of combustion is 4.0 × 10 ^{4} J/g ?**

**Ans.** Volume of water heated = 3.0 litre/minute \ Mass of water heated, m = 3000 g/minute Increase in temperature,

$$\Delta\text{T} = 77° \text{C} – 27° \text{C} = 50° \text{C}$$

Specific heat of water, c = 4.2 Jg^{–1}°C^{–1}

Amount of heat used, $$\text{Q} =\text{mc}\Delta \text{T}$$

or Q = 3000 g min^{–1} × 4.2 Jg–1 °C^{–1} × 50°C = 63 × 10^{4}J min^{–1}

Rate of combustion of fuel

$$= \frac{63×10^4\text{J}\space \text{min}^-1}{4.0×10^4\text{Jg}^-1}$$

=15.75 g min^{–1l}

Therefore, rate of consumption is 15.75 g/min

**12.2. 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 N**

_{2}= 28; R = 8.3 J**mol**

^{–1}K^{–1}.)**Ans.** Given,

Mass of nitrogen, m = 2 × 10–2 kg = 20 g

Rise in temperature, ΔT = 45°C

Heat required, Q = ?

Since, Q = nCT

We know,

$$C= \frac{7\text{R}}{2} (\text{diatomic\space molecule})\\\text{C}= \frac{7×8.3}{2}\\\text{Then, n(no. of moles)} =\frac {\text{m}}{\text{M}}(R=8.3\text{J/k-mol})$$

where,

Mass of gas(m) = 20 g

Molecular mass (M) = 28 u

$$\text{n}=\frac{20}{28}=\frac{10}{14}\text{moles}\\\text{n}=\frac{1}{1.4}\text{moles}$$

Let the temperature difference be 45 K.

$$\text{Then, Q}=\frac{10}{14}×\frac{7×8.3}{2}×45=\frac{8.3×225}{2}\\\therefore \text{Q}=933.75\space \text{J}$$

**12.3. Explain why **

**(a) Two bodies at different temperatures ****T _{1} and T_{2} if brought in thermal contact **

**do not necessarily settle to the mean**

**temperature (T**

_{1}+ T_{2})/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.**

**Ans.** (a) When two bodies with different temperatures, say T_{1} and T_{2}, are brought into thermal contact with each other, heat flows from the higher temperature body to the lower temperature body until both bodies attain an equilibrium position, i.e., both bodies acquire the same temperature. When the thermal capacity of both bodies are equal, the equilibrium temperature is equals to the mean temperature.

(b) A chemical or nuclear plant's coolant should always have a high specific heat. The coolant's capacity to absorb and release heat is proportional to its specific heat. As a result, the required coolant for a nuclear or chemical plant is a liquid with a high specific heat value. This would keep the plant from overheating in different parts.

(c) When a driver is driving a car, the air temperature inside the tyre rises due to the mobility of air molecules. By Charle's law, temperature is directly proportional to pressure.

Therefore, when the temperature inside a tyre increases, then there is also an increase of air pressure inside the tyre.

(d) The relative humidity in a harbour town is more than that of the relative humidity in a desert town. Humidity is a measure of water vapour in the atmosphere and the specific heat of water vapour is very high. Due to which temperature remains temperate or moderate. Therefore, the climate of a harbour town is more

temperate than that of a town in a desert at the same latitude.

**12.4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of 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 ?**

**Ans.** The cylinder is completely insulated from its surroundings.

Therefore, no heat is exchange takes place between the system (cylinder) and its surroundings.

Thus, the process is adiabatic in nature.

Initial pressure inside the cylinder = P1

Final pressure inside the cylinder = P2

Initial volume inside the cylinder = V1

Final volume inside the cylinder = V2

Ratio of specific heat,

$$\gamma =\frac{\text{C}_\text{p}}{\text{C}_\text{V}}=1.4$$

For an adiabatic process, we have:

$$\text{P}_1\text{V}_1\gamma =\text{P}_2\text{V}_2\gamma$$

The final volume is compressed to half of its initial volume

$$\text{i.e., V}_2=\space \frac{\text{V}_1}{2}\\\text{P}_1\text{V}_1\gamma = \text{P}_2(\text{V}_1/2)^\gamma\\\Rightarrow\frac{\text{P}_2}{\text{P}_1}=\frac{\text{V}_1^\gamma}{(\text{V}_1/2)^\gamma}\\=2^{1.4}\\or \frac{\text{P}_2}{\text{P}_1}= 2.639$$

Therefore, the pressure increases by a factor of 2.639.

**12.5. 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)**

**Ans.** Given :

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.

As given, it is an adiabatic process. Therefore, the change in heat is zero.

So, ΔQ = 0

Given, ΔW = – 22.3

(Since, the work is done on the system)

From first law of thermodynamics, we have:

ΔQ = ΔU + ΔW

Where, ΔU = Change in the internal

energy of the gas

Or DU = ΔQ – ΔW

= 0 – (– 22.3 J)

We get,

ΔU = + 22.3 J

When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

ΔQ = 9.35 cal

= 9.35 × 4.19 J

= 39.1765 J

and Heat absorbed,

ΔQ = ΔU + ΔW

Thus, ΔW = ΔQ – ΔU

= 39.1765 – 22.3

= 16.8765 J

Hence, 16.88 J of work is done by the system.

**12.6. 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 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 ?****Ans.** (a) As soon as the stopcock is opened, gas will begin to flow from cylinder A to cylinder B, which is totally evacuated. The volume of gas will be doubled because both cylinders have the same capacity. By Boyle's law pressure is inversely proportional to volume, the pressure will be reduced to half its initial value.

Since the gas in cylinder P has a starting pressure of 1 atm. As a result, each cylinder's pressure will now be 0.5 atm

$$\begin{pmatrix}\frac{1}{2}\text{atm} \end{pmatrix}$$

(b) The internal energy of the gas will not change in this situation, i.e. ΔU = 0. This is because of the fact that internal energy can only change when work is performed by or on the system. As no work is done by or on the gas in this situation. So internal energy will not change or remain unchanged.

(c) The temperature of the gas will remain unchanged. The reason for this is that the gas does not do any work while expanding. As a result, the temperature of the gas will not vary during this procedure.

(d) The given example is a clear case of free expansion, which is quick and uncontrollable. As the intermediate states do not meet the gas equation and are non-equilibrium, so they do not lie on the system's Pressure- Volume- Temperature surface.

**12.7. A steam engine delivers 5.4 × 10 ^{8 }**

**J of work**

**per minute and services 3.6 × 10**

^{9 }**J of heat per**

**minute from its boiler. What is the efficiency**

**of the engine? How much heat is wasted per**

**minute?**

**Ans.** Work done by the steam engine per minute,

W = 5.4 × 10^{8} J

Heat supplied from the boiler,

H = 3.6 × 10^{9}J

Efficiency of the engine,

$$\eta = \frac{\text{Output\space energy}}{\text{Input\space Energy}}\\\eta = \frac{\text{Work\space done}}{\text{Efficiency\space Energy\space Supplied}}\\= \frac {5.4×10^8}{3.6×10^9}\\=0.15\\=15\%$$

Therefore, the percentage efficiency of the engine is 15%.

Amount of heat energy wasted

$$= 3.6 × 10^9\\– 5.4 × 10^8\\= 30.6 × 10^8\text{ J}\\= 3.06 × 10^9\text{J}$$

Hence, the amount of heat wasted per minute is 3.06 ×10^{9} J.

**12.8. 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?****Ans.** According to law of conservation of energy

Total energy = work done + internal energy

ΔQ = ΔW + ΔU

Here,

Rate at which heat is supplied,

ΔQ = 100 W

Rate at which work is done

ΔW = 75 Js–1

Rate of change of internal energy is ΔU

ΔU = ΔQ – ΔW

ΔU = 100 – 75

= 25 J/s

Hence, the internal energy of the system is increasing at a rate of 25 J/s or 25 W.

**12.9. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13)**

**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.**

**Ans.** Total work done by the gas from D to E to F

= Area of ∆DEF

$$Area\space of\space \Delta \text{DEF}= \begin{pmatrix}\frac{1}{2} \end{pmatrix}×\text{DE}×\text{EF}\\\begin{pmatrix}\therefore \text{Area\space of\space triangle} = \frac{1}{2} ×\text{base}×\text{height}\end{pmatrix}\\$$

Where,

DF = Change in pressure

= 600 N/m^{2}

– 300 N/m^{2} = 300 N/

m^{2}

FE = Change in volume

= 5.0 m^{3}– 2.0 m^{3}

= 3.0 m^{3}

$$\therefore \text{Area\space of\space ∆\space DEF} = \begin{pmatrix}\frac{1}{2} \end{pmatrix}×300×3$$

= 450 J

Hence, the total work done by the gas from D to E to F is 450 J.

**12.10. A refrigerator is to maintain eatables kept inside at 9° C. If room temperature is 36° C, calculate the coefficient of performance.**

**Ans.** Temperature inside the refrigerator,

T^{1} = 9° C = (273 + 9) K = 282 K

Room temperature,

T^{2} = 36° C = (273 + 36) K = 309 K

$$\text{Coefficient\space of\space performance} = \frac{\text{T}_1}{\text{T}_1-\text{T}_1}$$

On substituting, we get,

$$= \frac{282}{(309-282)}$$

=10.44

Hence, the coefficient of performance of the given refrigerator is 10.44.

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