An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work is done be higher or lower than the earlier case? (Given that 1 L bar = 100J)
Amount of work done = -pext ∆V
= – 2 bar × (50 – 10 ) L = – 80 L bar.
Now, it is given that 1 L bar = 100 J
So , -80 L bar = (-80 × 100) = -8000 J
= -8 KJ is the amount of work done in the above process.
If it is reversible then the internal pressure will be larger than the external pressure. So the work done will own a higher value.
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