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.