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

Explanation

If m is the total mass of the gas then its kinetic energy =`1/2 mv^(2)`
When the vessel is suddenly stopped then total kinetic energy will increase the temperature of the gas. Hence
`1/2mv^(2)=muC_(v)DeltaT=m/MC_(v)DeltaT` [As `C_(v)=R/(gamma-1)`]
`rArrm/Mr/(gamma-1)DeltaT=1/2mv^(2)rArrDeltaT=(Mv^(2)(gamma-1))/(2R)`