Air pressure in a car tyre increases during driving. Explain.
During, driving, temperature of the gas increases while its volume remains constant.
So, according to Charle's law, at constant volume ($V$),
Pressure $(p) \propto$ Temperature $(T)$
Therefore, pressure of gas increases.
Consider a Carnot's cycle operating between $T_1=500 \mathrm{~K}$ and $T_2=300 \mathrm{~K}$ producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
$$\begin{aligned} \text{Given,}\quad \text { temperature of the source } T_1 & =500 \mathrm{~K} \\ \text { Temperature of the sink } T_2 & =300 \mathrm{~K} \\ \text { Work done per cycle } W & =1 \mathrm{~kJ}=1000 \mathrm{~J} \end{aligned}$$
Heat transferred to the engine per cycle $Q_1=$ ?
Efficiency of a Carnot engine $(\eta)=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}$
$$\text{and}\quad \eta=\frac{W}{Q_1}$$
$$\Rightarrow \quad Q_1=\frac{W}{\eta}=\frac{1000}{(2 / 5)}=2500 \mathrm{~J}$$
A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kcal , how many times must he go up and down to reduce his weight by 5 kg ?
Given, $\quad$ height of the stairs $=h=10 \mathrm{~m}$
Energy produced by burning 1 kg of fat $=7000 \mathrm{~kcal}$
$\therefore$ Energy produced by burning 5 kg of fat $=5 \times 7000=35000 \mathrm{~kcal}$ $=35 \times 10^6 \mathrm{~cal}$
$$\begin{aligned} &\text { Energy utilised in going up and down one time }\\ &\begin{aligned} & =m g h+\frac{1}{2} m g h=\frac{3}{2} m g h \\ & =\frac{3}{2} \times 60 \times 10 \times 10 \\ & =9000 \mathrm{~J}=\frac{9000}{4.2}=\frac{3000}{1.4} \mathrm{cal} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\therefore \text { Number of times, the person has to go up and down the stairs }\\ &\begin{aligned} & =\frac{35 \times 10^6}{(3000 / 1.4)}=\frac{35 \times 1.4 \times 10^6}{3000} \\ & =16.3 \times 10^3 \text { times } \end{aligned} \end{aligned}$$
Consider a cycle tyre being filled with air by a pump. Let $V$ be the volume of the tyre (fixed) and at each stroke of the pump $\Delta V(<< V)$ of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from $p_1$ to $p_2$ ?
Let, volume is increased by $\Delta V$ and pressure is increased by $\Delta p$ by an stroke. For just before and after an stroke, we can write
$$\begin{aligned} & p_1 V_1^\gamma =p_2 V_2^\gamma \\ \Rightarrow \quad & p(V+\Delta V)^\gamma =(p+\Delta p) V^\gamma \quad (\because \text { volume is fixed })\\ \Rightarrow \quad & p V^\gamma\left(1+\frac{\Delta V}{V}\right)^\gamma =p\left(1+\frac{\Delta p}{p}\right) V^\gamma \\ \Rightarrow \quad & p V^\gamma\left(1+\gamma \frac{\Delta V}{V}\right) & \approx p V^\gamma\left(1+\frac{\Delta p}{p}\right) \quad (\because \Delta v < < v)\\ \Rightarrow \quad & \gamma \frac{\Delta V}{V} =\frac{\Delta p}{p} \Rightarrow \Delta V=\frac{1}{\gamma} \frac{V}{p} \Delta p \\ \Rightarrow \quad & d V =\frac{1}{\gamma} \frac{V}{p} d p \end{aligned}$$
$$\begin{aligned} &\text { Hence, work done is increasing the pressure from } p_1 \text { to } p_2\\ &\begin{aligned} W & =\int_{p_1}^{p_2} p d V=\int_{p_1}^{p_2} p \times \frac{1}{\gamma} \frac{V}{p} d p \\ & =\frac{V}{\gamma} \int_{p_1}^{p_2} d p=\frac{V}{\gamma}\left(p_2-p_1\right) \\ \Rightarrow \quad W & =\frac{\left(p_2-p_1\right)}{\gamma} V \end{aligned} \end{aligned}$$
In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power and heat transferred from $-3^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$, find the heat taken out of the refrigerator per second assuming its efficiency is $50 \%$ of a perfect engine.
Given, temperature of the source is $27^{\circ} \mathrm{C}$
$$\begin{array}{ll} \Rightarrow & T_1=(27+273) \mathrm{K}=300 \mathrm{~K} \\ \text { Temperature of sink } & T_2=(-3+273) \mathrm{K}=270 \mathrm{~K} \end{array}$$
Efficiency of a perfect heat engine is given by
$$\eta=1-\frac{T_2}{T_1}=1-\frac{270}{300}=\frac{1}{10}$$
Efficiency of refrigerator is $50 \%$ of a perfect engine
$$\therefore \quad \eta^{\prime}=0.5 \times \eta=\frac{1}{2} \eta=\frac{1}{20}$$
$$\begin{aligned} &\therefore \text { Coefficient of performance of the refrigerator }\\ &\begin{aligned} \beta & =\frac{Q_2}{W}=\frac{1-\eta^{\prime}}{\eta^{\prime}} \\ & =\frac{1-(1 / 20)}{(1 / 20)}=\frac{19 / 20}{1 / 20}=19 \end{aligned} \end{aligned}$$
$$\Rightarrow \quad Q_2=\beta W=19 W \quad\left(\because \beta=\frac{Q_2}{W}\right)$$
$$\begin{aligned} &=19 \times(1 \mathrm{~kW})=19 \mathrm{~kW}=19 \mathrm{~kJ} / \mathrm{s} .\\ &\text { Therefore, heat is taken out of the refrigerator at a rate of } 19 \mathrm{~kJ} \text { per second. } \end{aligned}$$