A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle. $\left[C_V=(3 / 2) R\right]$
(a) AB : constant volume
(b) BC : constant pressure
(c) CD : adiabatic
(d) DA : constant pressure
(a) For process $A B$,
Volume is constant, hence work done $d W=0$
Now, by first law of thermodynamics,
$$\begin{aligned} & d Q=d U+d W=d U+0=d U \\ & =n C V d T=n C V\left(T_B-T_A\right) \\ & =\frac{3}{2} R\left(T_B-T_A\right) \quad (\because n=1)\\ & =\frac{3}{2}\left(R T_B-R T_A\right)=\frac{3}{2}\left(p_B V_B-p_A V_A\right) \\ & \text { Heat exchanged }=\frac{3}{2}\left(p_B V_B-p_A V_A\right) \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { (b) For process BC,}\quad p & =\text { constant } \\ d Q & =d U+d W=\frac{3}{2} R\left(T_C-T_B\right)+p_B\left(V_C-V_B\right) \\ & =\frac{3}{2}\left(p_C V_C-p_B V_B\right)+p_B\left(V_C-V_B\right)=\frac{5}{2} p_B\left(V_C-V_B\right) \end{aligned} \end{aligned}$$
Heat exchanged $=\frac{5}{2} p_B\left(V_C-V_A\right) \quad \left(\because p_B=p_C\right.$ and $\left.p_B=V_A\right)$
(c) For process $C D$, Because $C D$ is adiabatic, $d Q=$ Heat exchanged $=0$
(d) $D A$ involves compression of gas from $V_D$ to $V_A$ at constant pressure $p_A$.
$\therefore$ Heat exchanged can be calculated by similar way as $B C_1$,
Hence, $$d Q=\frac{5}{2} p_A\left(V_A-V_D\right)$$
Consider that an ideal gas ( $n$ moles) is expanding in a process given by $p=f(V)$, which passes through a point $\left(V_0, p_0\right)$. Show that the gas is absorbing heat at $\left(p_0, V_0\right)$ if the slope of the curve $p=f(V)$ is larger than the slope of the adiabatic passing through $\left(p_0, V_0\right)$.
According to question, slope of the curve $=f(V)$, where $V$ is volume .
$$\begin{array}{rr} \therefore \quad & \text { Slope of } p=f(V) \text { curve at }\left(V_0, p_0\right)=f\left(V_0\right) \\ & \text { Slope of adiabatic at }\left(V_0, p_0\right)=k(-\gamma) V_0^{-1-\gamma}=-\gamma p_0 / V_0 \end{array}$$
Now heat absorbed in the process $p=f(V)$
$$d Q=d U+d W=n C_V d T+p d V\quad \text{... (i)}$$
$$\because \quad p V=n R T \Rightarrow T=\left(\frac{1}{n R}\right) p V$$
$$\begin{array}{ll} \Rightarrow & T=\left(\frac{1}{n R}\right) V f(V) \\ \Rightarrow & d T=\left(\frac{1}{n R}\right)\left[f(V)+V f^{\prime}(V)\right] d V\quad \text{... (ii)} \end{array}$$
$$\begin{aligned} &\text { Now from Eq. (i) }\\ &\begin{aligned} \frac{d Q}{d V} & =n C_V \frac{d T}{d V}+p \frac{d V}{d V}=n C_V \frac{d T}{d V}+p \\ & =\frac{n C_V}{n R} \times\left[f(V)+V f^{\prime}(V)\right]+p \quad \text{[from Eq. (ii)]}\\ & =\frac{C_V}{R}\left[f(V)+V f^{\prime}(V)\right]+f(V) \quad [\because p=f(V)] \end{aligned} \end{aligned}$$
$$\left.\Rightarrow \quad\left[\frac{d Q}{d V}\right]_{V=V_0}=\frac{C_V}{R}\left[f\left(V_0\right)+V_0 f^{\prime}\right]\left(V_0\right)\right]+f\left(V_0\right)$$
$$=f\left(V_0\right)\left[\frac{C_V}{R}+1\right]+V_0 f^{\prime}\left(V_0\right) \frac{C_V}{R}$$
$$\begin{aligned} \because \quad C_V & =\frac{R}{\gamma-1} \Rightarrow \frac{C_V}{R}=\frac{1}{\gamma-1} \\ \Rightarrow \quad\left[\frac{d Q}{d v}\right]_{V=V_0} & =\left[\frac{1}{\gamma-1}+1\right] f\left(V_0\right)+\frac{V_0 f^{\prime}\left(V_0\right)}{\gamma-1} \\ & =\frac{\gamma}{\gamma-1} p_0+\frac{V_0}{\gamma-1} f^{\prime}\left(V_0\right) \end{aligned}$$
Heat is absorbed where $\frac{d Q}{d V}>0$, when gas expands
Hence, $\gamma p_0+V_0 f^{\prime}\left(V_0\right)>0 \quad$ or $f^{\prime}\left(V_0\right)>\left(-\gamma \frac{p_0}{V_0}\right)$
Consider one mole of perfect gas in a cylinder of unit cross-section with a piston attached (figure). A spring (spring constant $k$ ) is attached (unstretched length $L$ ) to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat $Q$ is supplied to the gas causing an increase of value from $V_0$ to $V_1$.
(a) What is the initial pressure of the system?
(b) What is the final pressure of the system?
(c) Using the first law of thermodynamics, write down a relation between $Q, p_a, V, V_0$ and $k$.
$$\text { (a) Initially the piston is in equilibrium hence, } p_i=p_a$$
(b) On supplying heat, the gas expands from $V_0$ to $V_1$
$\therefore$ Increase in volume of the gas $=V_1-V_0$
As the piston is of unit cross-sectional area hence, extension in the spring
$$x=\frac{V_1-V_0}{\text { Area }}=V_1-V_0 \quad \text{[Area=1]}$$
$\therefore$ Force exerted by the spring on the piston
$$=F=k x=k\left(V_1-V_0\right)$$
Hence,
$$\begin{aligned} \text { final pressure } & =p_f=p_a+k x \\ & =p_a+k \times\left(V_1-V_0\right) \end{aligned}$$
(c) From first law of thermodynamics $d Q=d U+d W$
If $T$ is final temperature of the gas, then increase in internal energy
$$\begin{aligned} d U & =C_V\left(T,-T_0\right)=C_V\left(T,-T_0\right) \\ \text{We can write,} \quad T & =\frac{p_t V_1}{R}=\left[\frac{p_a+k\left(V_1-V_0\right)}{R}\right] \frac{V_1}{R} \end{aligned}$$
Work done by the gas $=p d V+$ increase in PE of the spring
$$=p_a\left(V_1-V_0\right)+\frac{1}{2} k x^2$$
Now, we can write $d Q=d U+d W$
$$\begin{aligned} & =C_V\left(T-T_0\right)+p_a\left(V-V_0\right)+\frac{1}{2} k x^2 \\ & =C_V\left(T-T_0\right)+p_a\left(V-V_0\right)+\frac{1}{2}\left(V_1-V_0\right)^2 \end{aligned}$$
This is the required relation.