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.
A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in figure.
$A$ to $B$ volume constant, $B$ to $C$ adiabatic, $C$ to $D$ volume constant and $D$ to $A$ adiabatic
$$V_C=V_D=2 V_A=2 V_B$$
(a) In which part of the cycle heat is supplied to the engine from outside?
(b) In which part of the cycle heat is being given to the surrounding by the engine?
(c) What is the work done by the engine in one cycle? Write your answer in term of $p_A, p_B, V_A$ ?
(d) What is the efficiency of the engine? ( $\gamma=\frac{5}{3}$ for the gas), $\left(C_V=\frac{3}{2} R\right.$ for one mole)
For the process AB,
$$\begin{aligned} &\begin{aligned} & d V=0 \Rightarrow d W=0 \quad (\because \text { volume is constant) }\\ & d Q=d U+d W=d U \\ \Rightarrow \quad & d Q=d U=\text { Change in internal energy. } \end{aligned}\\ \end{aligned}$$
Hence, in this process heat supplied is utilised to increase, internal energy of the system.
Since, $p=\left(\frac{n R}{V}\right) T$, in isochoric process, $T \propto p$. So temperature increases with increases of pressure in process $A B$ which inturn increases internal energy of the system i.e., $d U>0$. This imply that $d Q>0$. So heat is supplied to the system in process $A B$.
(b) For the process $C D$, volume is constant but pressure decreases.
Hence, temperature also decreases so heat is given to surroundings.
(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.
$$\begin{aligned} W_{A B} & =\int_A^B p d V=0, W_{C D}=\int_{V_C}^{V_D} p d V=0 \quad (\because dV=0)\\ W_{B C} & =\int_{V_B}^{V_C} p d V=k \int_{V_B}^{V_C} \frac{d V}{V^\gamma}=\frac{k}{1-\gamma}\left[V^{1-\gamma}\right]_{V_B}^{V_C} \\ & =\frac{1}{1-\gamma}[p V]_{V_B}^{V_C}=\frac{\left(p_C V_C-p_B V_B\right)}{1-\gamma} \\ \text{Similarly,}\quad W_{D A} & =\frac{p_A V_A-p_D V_D}{1-\gamma} \quad [\because \text{BC is adiabatic process}] \end{aligned}$$
$$\because$$ B and C lies on adiabatic curve BC.
$$\begin{aligned} \therefore \quad p_B V_B^\gamma & =p_C V_C^\gamma \\ p_C & =p_B\left(\frac{V_B}{V_C}\right)^\gamma=p_B\left(\frac{1}{2}\right)^\gamma=2^{-\gamma} p_B \end{aligned}$$
Similarly, $$p_D=2^{-\gamma} p_A$$
$$\begin{aligned} &\text { Total work done by the engine in one cycle } A B C D A \text {. }\\ &\begin{aligned} W & =W_{A B}+W_{B C}+W_{C D}+W_{D A}=W_{B C}+W_{D A} \\ & =\frac{\left(p_C V_C-p_B V_B\right)}{1-\gamma}+\frac{\left(p_A V_A-p_D V_D\right)}{1-\gamma} \\ W & =\frac{1}{1-\gamma}\left[2^{-\gamma} p_B\left(2 V_B\right)-p_B V_B+p_A V_A-2^{-\gamma} p_B\left(2 V_B\right)\right] \\ & =\frac{1}{1-\gamma}\left[p_B V_B\left(2^{-\gamma+1}-1\right)-p_A V_A\left(2^{-\gamma+1}-1\right)\right. \\ & =\frac{1}{1-\gamma}\left(2^{1-\gamma}-1\right)\left(p_B-p_A\right) V_A \\ & =\frac{3}{2}\left[1-\left(\frac{1}{2}\right)^{2 / 3}\right]\left(p_B-p_A\right) V_A \end{aligned} \end{aligned}$$