An insulated box containing a monoatomic gas of molar mass (M) moving with a speed `v_0` is suddenly stopped. Find the increment is gas temperature as a result of stopping the box.

Options

A. `(Mv^(2))/(3R)`
B. `(Mv^(2))/(5R)`
C. `(2Mv^(2))/(5R)`
D. `(Mv^(2))/(5R)`

Answer

B

Explanation

The K.E. of the gas = `n(1/2Mv^(2))" "....(1)`
where n = no. of moles of the gas in the box.
When the box stops suddenly, the K.E. of the gas increases the internal energy and hence the temperature of the gas.
`DeltaU=nC_(V)dT=nxx5/2RxxdT" "...(2)`
where `C_(V)=(5R)/2` for a diatomic gas
Equating (1) and (2), we get
`(1/2Mv^(2))n=n(5/2R)dT`
`thereforedT=(Mv^(2))/(5R)`