Question

An ideal gas occupying a volume of $2$ $d{m}^3$ and pressure of $5$ bar undergoes isothermal and irreversible expansion against an external pressure of $1$ bar.

Find the work done in the process?

Answer 1

Solution:-

As we know that, in an isothemal expansion,

$PV=\text{Constant}$

$\frac{P_i}{P_f}=\frac{V_f}{V_i}$

$\Rightarrow V_f=\frac{P_i\times V_i}{P_f}$

Given that,

$P_i=5\text{bar}$,$P_f=1\text{bar}$ and $V_i=2{dm}^3$

$\therefore V_f=\frac{5\times 2}{1}=10{dm}^3$

Now, work done $\left(W\right)=-P_{ext.}(V_f-V_i)$

$\therefore W=-1(10-2)$

$\Rightarrow W=-8bar-{dm}^3=-800J$

Hence the final volume of the system is $10{dm}^3$ and the work involved in the process is $-800J$.