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
$$\therefore \quad \eta^{\prime}=0.5 \times \eta=\frac{1}{2} \eta=\frac{1}{20}$$
$$\begin{aligned} &\therefore \text { Coefficient of performance of the refrigerator }\\ &\begin{aligned} \beta & =\frac{Q_2}{W}=\frac{1-\eta^{\prime}}{\eta^{\prime}} \\ & =\frac{1-(1 / 20)}{(1 / 20)}=\frac{19 / 20}{1 / 20}=19 \end{aligned} \end{aligned}$$
$$\Rightarrow \quad Q_2=\beta W=19 W \quad\left(\because \beta=\frac{Q_2}{W}\right)$$
$$\begin{aligned} &=19 \times(1 \mathrm{~kW})=19 \mathrm{~kW}=19 \mathrm{~kJ} / \mathrm{s} .\\ &\text { Therefore, heat is taken out of the refrigerator at a rate of } 19 \mathrm{~kJ} \text { per second. } \end{aligned}$$
If the coefficient of performance of a refrigerator is 5 and operates at the room temperature $\left(27^{\circ} \mathrm{C}\right)$, find the temperature inside the refrigerator.
$$\begin{aligned} \text { Given, coefficient of performace }(\beta) & =5 \\ T_1 & =(27+273) \mathrm{K}=300 \mathrm{~K}, T_2=\text { ? } \end{aligned}$$
$$\begin{aligned} \text { Coefficient of performance }(\beta) & =\frac{T_2}{T_1-T_2} \\ 5 & =\frac{T_2}{300-T_2} \Rightarrow 1500-5 T_2=T_2 \end{aligned}$$
$$\begin{array}{l} \Rightarrow \quad & 6 T_2=1500 \quad \Rightarrow \quad T_2=250 \mathrm{~K} \\ \Rightarrow \quad & T_2 =(250-273)^{\circ} \mathrm{C}=-23^{\circ} \mathrm{C} \end{array}$$
The initial state of a certain gas is ( $p_i, V_i, T_i$ ). It undergoes expansion till its volume becomes $V_f$. Consider the following two cases
(a) the expansion takes place at constant temperature.
(b) the expansion takes place at constant pressure.
Plot the $p-V$ diagram for each case. In which of the two cases, is the work done by the gas more?
Consider the diagram p-V, where variation is shown for each process.
Process 1 is isobaric and process 2 is isothermal. Since, work done $=$ area under the $p-V$ curve. Here, area under the $p-V$ curve 1 is more. So, work done is more when the gas expands in isobaric process.
Consider a $p-V$ diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure.
(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature $T_1 / T_2$, if $V_2=2 V_1$ ?
(c) Given the internal energy for one mole of gas at temperature $T$ is (3/2)RT, find the heat supplied to the gas when it is taken from state 1 to 2 , with $V_2=2 V_1$.
Let $p V^{1 / 2}=$ Constant $=K, p=\frac{K}{\sqrt{V}}$
(a) Work done for the process 1 to 2 ,
$$\begin{aligned} W & =\int_{V_1}^{V_2} p d V=K \int_{V_1}^{V_2} \frac{d V}{\sqrt{V}}=K\left[\frac{\sqrt{V}}{1 / 2}\right]_{V_1}^{V_2}=2 K\left(\sqrt{V_2}-\sqrt{V_1}\right) \\ & =2 p_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right)=2 p_2 V_2^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right) \end{aligned}$$
(b) From ideal gas equation,$$\begin{aligned} &\begin{aligned} p V & =n R T \quad \Rightarrow \quad T=\frac{p V}{n R}=\frac{p \sqrt{V} \sqrt{V}}{n R} \\ T & =\frac{K \sqrt{V}}{n R} \quad (\mathrm{As}, p \sqrt{V}=K) \end{aligned}\\ \end{aligned}$$
$$\text{Hence,}\quad T_1=\frac{K \sqrt{V_1}}{n R} \Rightarrow T_2=\frac{K \sqrt{V_2}}{n R}$$
$$\Rightarrow \quad \frac{T_1}{T_2}=\frac{\frac{K \sqrt{V_1}}{n R}}{\frac{K \sqrt{V_2}}{n R}}=\sqrt{\frac{V_1}{V_2}}=\sqrt{\frac{V_1}{2 V_1}}=\frac{1}{\sqrt{2}} \quad\left(\because V_2=2 V_1\right)$$
$$\begin{aligned} &\text { (c) Given, internal energy of the gas }=U=\left(\frac{3}{2}\right) R T\\ &\begin{aligned} \Delta U & =U_2-U_1=\frac{3}{2} R\left(T_2-T_1\right) \\ & =\frac{3}{2} R T_1(\sqrt{V}-1) \end{aligned}\\ &\left[\because T_2=\sqrt{2} T_1 \text { from }(\mathrm{b})\right] \end{aligned}$$
$$ \begin{aligned} \Delta W & =2 p_1 V_1^{1 / 2}\left(\sqrt{V_2}-\sqrt{V_1}\right) \\ & =2 p_1 V_1^{1 / 2}\left(\sqrt{2} \times \sqrt{V_1}-\sqrt{V_1}\right) \\ & =2 p_1 V_1(\sqrt{2}-1)=2 R T_1(\sqrt{2}-1) \\ \therefore & \quad \Delta Q=\Delta U+\Delta W \\ & =\frac{3}{2} R T_1(\sqrt{2}-1)+2 R T_1(\sqrt{2}-1) \\ & =(\sqrt{2}-1) R T_1(2+3 / 2) \\ & =\left(\frac{7}{2}\right) R T_1(\sqrt{2}-1) \end{aligned}$$
This is the amount of heat supplied.