An insulated container containing monoatomic gas of molar mass m is moving with a velocity `V_(0)`. If the container is suddenly stopped , find the change in temperature .

Options

A. `(mv_(0)^(2))/(2R)`
B. `(mv_(0)^(2))/(3R)`
C. `(R )/(mv_(0)^(2))`
D. `(3mv_(0)^(2))/(2R)`

Answer

B

Explanation

If the container is suddenly stopped loss in kinetic energy of gas `=1/2(mn)v_(0)^(2),` where n is number of moles of gas.
Let `DeltaT` is the fall in temperature of gas,
Then,`n((3)/(2)RDeltaT)=1/2mnv_(0)^(2)`
`DeltaT=(mv_(0)^(2))/(3R)`