A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats `gamma`. It is moving with speed v and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:

Options

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

Answer

D

Explanation

The K.E. of the vessel = `1/2mv^(2)`
When it is suddenly stopped, its K.E. is used to increase the temperature of the gas.
The increase in internal energy = `DeltaU=nC_(v)dT` where `C_(v)` is the specific heat of the gas.
`therefore1/2mv^(2)=nC_(v)dT`
But `n=m/MandC_(v)=R/(gamma-1)`
`therefore1/2mv^(2)=m/M*(R/(gamma-1))*dT`
`thereforedT=(Mv^(2)(gamma-1))/(2R)`