Question

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m3m3 and the pressure increases from 2.50×105 PaPa to 5.50×105 PaPa . The second process is a compression to a volume of 0.110 m3m3 at a constant pressure of 5.50×105 PaPa.

Answer 1

`W = -4.95 * 10^4\ J`

Step-by-step explanation:

given,

In first case Volume remains constant.

Work done in the first case is zero.

In Second case Volume change

V₁ = 0.2 m³

V₂ = 0.11 m³

Pressure, P = 5.5 x 10⁵ Pa

Work done = Pressure x change in volume

W = P ΔV

`W = P(V_2-V_1)`

`W = 5.5*10^5* (0.11 - 0.2)`

`W = -4.95 * 10^4\ J`

Hence, Work done when volume changes is equal to
`W = -4.95 * 10^4\ J`