A fixed mass of helium gas is made to undergo a process in which its pressure varies linearly from `1 kPa` to `2 kPa`, in relation to its volume as the latter varies from `0.2 m^(3)` to `0.4 m^(3)`. The heat absorbed by the gas will be

Options

A. `300 J`
B. `900 J`
C. `1200 J`
D. `1500 J`

Answer

C

Explanation

`Delta W = 1/2(1+2)xx(10^3N)/(m^2)xx(0.4-0.2)m^(3) = 300J`

`Delta U = nC_(V)DeltaT = ([P_(2)V_(2)-P_(1)V_(1)]/(gamma-1)`
`=(3//2)[0.4xx2-0.2xx1]xx10^(3)J = 900J`
`DeltaQ = Delta W + Delta U = 1200J`.