Certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velcity `v`. The molecular mass of gas is `M`. The rise in temperature of the gas when the vessel is suddenly stopped is `(gamma C_(P) // C_(V))`

Options

A. `(Mv^(2) (gamma -1))/(2 R (gamma + 1))`
B. `(Mv^(2) (gamma -1))/(2 R)`
C. `(Mv^(2))/(2 R (gamma + 1))`
D. `(Mv^(2))/(2 R (gamma - 1))`

Answer

B

Explanation

b. If `m` is the mass of the gas, then its kinetic energy
`= 1//2 mv^(2)`
When the vessel is suddenly stopped, total kinetic energy will increase the temperature of the gas (because process will be adiabatic), i.e.,
`(1)/(2) mv^(2) = mu C_(v) Delta T`
`= (m)/(M) C_(v) Delta T`
`implies (m)/(M) (R )/(gamma - 1) Delta T = (1)/(2) mv^(2)` `(As C_(v) = (R )/(gamma- 1))`
`implies Delta T = (M v^(2) (gamma - 1))/(2R)`