The ratio of specific heats of a gas is `gamma`. The change in internal energy of one mole of the gas, when the volume change from V to 2V at constant pressure p is

Options

A. `(gamma-1)/(pV)`
B. pV
C. `(pV)/(gamma -1)`
D. `(pV)/(gamma)`

Answer

C

Explanation

From first law of thermodynamics,
dQ = dU + dW
`implies dU = dQ - dW `
At constant pressure , `dQ = C_(p)dT, dW = pdV`
` dU = C_(p)dT - pdV`
`= C_(p) dT - RdT`
= ` (C_(p)-R)dT = C_(v) dT`
` (R)/(gamma - 1) dT = (R)/(gamma-1) xx (pV)/(R) = (pV)/(gamma-1)`
`therefore du = (pV)/(gamma -1)`