Question

One mole of an ideal gas is heated slowly so that it goes from the $P-V$ state $(P_i,V_i)$ to $(3P_i,3V_i)$ in such a way that the pressure of the gas is directly proportional to the volume. How much work is done on the gas in the process?

• A. $4P_i{V}_i$
• B. $16P_i{V}_i$
• C. $-9P_i{V}_i$
• D. $-4P_i{V}_i$

Answer 1

The correct option is D $-4P_i{V}_i$

Given $P\propto V$
i.e $P=kV$
So, during the heating process, $P=\left(\frac{P_i}{V_i}\right)V$
[since $P_i=k{V}_i$]
Work done on the gas is
$W=-{\int }_{V_i}^{V_f}PdV=-{\int }_{V_i}^{3V_i}\left(\frac{P_i}{V_i}\right)VdV$
$W=-\left(\frac{P_i}{V_i}\right){\left[\frac{V^2}{2}\right]}_{V_i}^{3V_i}=-\frac{P_i}{2V_i}(9V_i^2-V_i^2)$
$W=-4P_i{V}_i$