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. `(3mv_(0)^(2))/(2R)`
D. `(1)/(2)(mv_(0)^(2))/(R)`

Answer

B

Explanation

Here, M= mole mass of the gas `v_(o)` = initial speed
Mass of gas = n m (n= number of moles in the gas) `therefore` Initial Kinetic energy of the gas `=1/2(nm)v^(2)` Final KE of gas = 0
Change in Kinetic energt, `DeltaK=1/2(nm)v^(2)`
Let change in temperature of gas = `DeltaT`
Change in internal energy of the gas
`DeltaU=nC_(v)DeltaT=n((3)/(2)R)DeltaT`
As, `DeltaU=DeltaK`
`implies3/2nRDeltaT=1/2mnv_(o)^(2)`
`impliesDeltaT=(mv_(0)^(2))/(3R)`