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

Answer:

$W = -4.95 \times 10^4\ J$

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

$W = P(V_2-V_1)$

$W = 5.5\times10^5\times (0.11 - 0.2)$

$W = -4.95 \times 10^4\ J$

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