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.

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 .

overcharge is the right answer system