NCERT Solutions for Class 11 Physics Chapter 12 Thermodynamics

Ans : \(\begin{array}{l}{\text { Water is flowing at a rate of } 3.0 \text { litre/min. }} \\ {\text { The geyser heats the water, raising the temperature from } 27^{\circ} \mathrm{C} \text { to } 77^{\circ} \mathrm{C} \text { . }} \\ {\text { Initial temperature, } T_{1}=27^{\circ} \mathrm{C}} \\ {\text { Final temperature, } T_{2}=77^{\circ} \mathrm{C}} \\ {\therefore \text { Rise in temperature, } \Delta T=T_{2}-T_{1}} \\ {=77-27=50^{\circ} \mathrm{C}}\end{array}\)
\(\begin{array}{l}{\text { Heat of combustion }=4 \times 10^{4} \mathrm{Jg}} \\ {\text { Specific heat of water, } c=4.2 \mathrm{g}^{-1} \mathrm{c}^{-1}} \\ {\text { Mass of flowing water, } m=3.0 \text { litre/min }=3000 \mathrm{g} / \mathrm{min}} \\ {\text { Total heat used, } \Delta Q=m c \Delta T} \\ {=3000 \times 4.2 \times 50} \\ {=6.3 \times 10^{5} \mathrm{J} / \mathrm{min}}\end{array}\)
\(\therefore \text { Rate of consumption }=\frac{6.3 \times 10^{5}}{4 \times 10^{4}}=15.75 \mathrm{g} / \mathrm{min}\) 

Ans : \(\begin{array}{l}{\text { Mass of nitrogen, } m=2.0 \times 10^{-2} \mathrm{kg}=20 \mathrm{g}} \\ {\text { Rise in temperature, } \Delta T=45^{\circ} \mathrm{C}} \\ {\text { Molecular mass of } \mathrm{N}_{2}, \mathrm{M}=28} \\ {\text { Universal gas constant, } R=8.3 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}}\end{array}\)
\(\begin{array}{l}{n=\frac{m}{M}} \\ {\text { Number of moles, }^{n=\frac{m}{M}}} \\ {=\frac{2.0 \times 10^{-2} \times 10^{3}}{28}=0.714}\end{array}\)
\(\begin{array}{l}{\text { Molar specific heat at constant pressure for nitrogen, }^{C_{p}=\frac{7}{2} R}} \\ {=\frac{7}{2} \times 8.3} \\ {\text { The total amount of heat to be supplied is given by the relation: }} \\ {\Delta Q=n C_{P} \Delta T}\end{array}\)
\(\begin{array}{l}{=0.714 \times 29.05 \times 45} \\ {=933.38 \mathrm{J}} \\ {\text { Therefore, the amount of heat to be supplied is } 933.38 \mathrm{J} \text { . }}\end{array}\) 

Ans : (a) When two bodies at different temperatures Tl and T2 are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures Of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T1+T2)/2 only when the thermal capacities of both the bodies are equal. 
(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a hi$l specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot. 
(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles' law, temperature is directly 
proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase. 
(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town. 

Ans : The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic. 
\(\begin{array}{l}{\text { Initial pressure inside the cylinder }=P_{1}} \\ {\text { Final pressure inside the cylinder }=P_{2}} \\ {\text { Initial volume inside the cylinder }=V_{1}} \\ {\text { Final volume inside the cylinder }=V_{2}} \\ {\text { Ratio of specific heats, } Y=1.4} \\ {\text { For an adiabatic process, we have: }} \\ {\quad P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}}\end{array}\)
The final volume is compressed to half of its initial volume. 
\(V_{2}=\frac{V_{1}}{2}\)
\(\begin{array}{l}{P_{1}\left(V_{1}\right)^{\gamma}=P_{2}\left(\frac{V_{1}}{2}\right)^{\gamma}} \\ {\frac{P_{2}}{P_{1}}=\frac{\left(V_{1}\right)^{\gamma}}{\left(\frac{V_{1}}{2}\right)^{T}}=(2)^{\gamma}=(2)^{1,4}=2.639} \\ {\text { Hence, the pressure increases by a factor of } 2.639 .}\end{array}\) 

Ans : \(\begin{array}{l}{\text { The work done (W) on the system while the gas changes from state } A \text { to state } B \text { is } 22.3} \\ {\text { J. }} \\ {\text { This is an adiabatic process. Hence, change in heat is zero. }} \\ {\therefore \Delta Q=0} \\ {\Delta W=-22.33 \text { (Since the work is done on the system) }} \\ {\text { From the first law of thermodynamics, we have: }}\end{array}\)
\(\begin{array}{l}{\Delta Q=\Delta U+\Delta W} \\ {\text { Where, }} \\ {\Delta U=\text { Change in the internal energy of the gas }} \\ {\therefore \Delta U=\Delta Q-\Delta W=-(-22.3]} \\ {\Delta U=+22.3 \mathrm{J}}\end{array}\)
\(\begin{array}{l}{\text { When the gas goes from state } A \text { to state } B \text { via a process, the net heat absorbed by the }} \\ {\text { system is: }} \\ {\Delta Q=9.35 \text { cal }=9.35 \times 4.19=39.1765 \mathrm{J}} \\ {\text { Heat absorbed, } \Delta Q=\Delta U+\Delta Q} \\ {\therefore \Delta W=\Delta Q-\Delta U} \\ {=39.1765-22.3} \\ {=16.8765 J}\end{array}\)