An ideal gas is allowed to expand against a constant pressure of $2bar$ from $10L$ to $50L$ in one step.

A. Calculate the amount of work done by the gas. (Given that $1Lbar=100J)$

B. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case?

Here, one needs to compare the work done in expanding an ideal gas from $10L$ to $50L$ in one step or finite step with the work done by the same gas in an infinite step or in reversible manner.

$\text{Calculating work done by gas}$

Given data:

$P=2bar$
$V_1=initialvolume=10L$
$V_2=finalvolume=50L$

The process is carried out in one step, i.e. finite step. So, it is irreversible process

$\therefore W=-p_{ext}\left(\Delta V\right)$

$w=workdonebysystemduringexpansion$

$p_{ext}=changeinvolume$

$\Delta V=Changeinvolume$

$\therefore W=-2bar(V_2–V_1)$

$=-2bar(50-10)L$ $=-80Lbar$

$(1Lbar=100J$
$80Lbar=8000J)$

$\therefore W=-8000J$ or $–8kJ$

The negative sign shows that work is done by the system on the surroundings, during expansion.


$\text{B. Work done comparison in reversible and irreversible process}$

Work done will be more in the reversible expansion because internal pressure and external pressure are almost same at every step or they vary infinitesimally.

i.e. Maximum work is obtained in reversible process than in irreversible process during expansion.

Hence the given problem is solved

i) Work done by the system during irreversible (one step) process $=-8KJ$

ii) But, more work is obtained in reversible (infinite steps) process during expansion.

This can also be confirmed from area under the $PV$ curve, that for reversible process maximum work is obtained in reversible way than irreversible during expansion.