A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3 and the pressure increases from 1.00*105 Pa to 4.00*105 Pa. The second process is a compression to a volume of 5.00*10-2 m3 at a constant pressure of 4.00*105 Pa. Find the total work done by the gas during both processes.

The work done = - 2 x 10⁴ J

Explanation:

In the first case, the volume is kept constant and pressure varies.

In isothermal process, the work done

W₁ = V x ΔP

here V is the volume of gas and ΔP is the change in pressure

Thus W₁ = 0

Because there is no change in volume, therefore displacement is zero.

In second case pressure is constant, but volume changes

Thus W₂ = P x ΔV

here P is the pressure and ΔV is the change in volume

Therefore W₂ = 4 x 10⁵ x 5 x 10⁻² = 2 x 10⁴ J

The total work done W = - 2 x 10⁴ J

Because the work done in compression is negative.