The internal energy of a gas is given by `U=1.5pV`. It expands from `100 cm^(3) to 200 cm^(3)` against a constant pressure of `1.0xx10^(5)` Pa. calculate the heat absorbed by the gas in the process.

Answer

B

Explanation

Delta U = 1.5pV`
`DeltaV = (20-100)cm^3 = 100cm^3`
` = 10^(-4)m^3`
`P = 1xx10^5pa`
`Delta U = 1.5xx10^5xx10^(-4) = 15J`
`Delta W = 10^5xx10^(-4) = 10J`
`DeltaQ = Delta U+DeltaW = 10+15 = 25J`