The mean kinetic energy of monoatomic gas molecules under standard conditions is `(E_(1))`. If the gas is compressed adiabatically 8 times to its initial volume, the mean kinetic energy of gas molecules changes to `(E_(2))`. The ratio `((E_(2)))/((E_(1)))` is

Options

A. 2
B. 4
C. 6
D. 8

Answer

B

Explanation

In adiabatic process, `TV^(gamma-1)=` constant
In given conditions,
`V_(1) = V`(say), so `V_(2) = ((V)/(8))`
and `gamma = (5)/(3)` (gas is monoatom)
So, `T_(1) V^((5)/(3) -1) = T_(2) ((V)/(8))^((5)/(3)-1) rArr (T_(2))/(T_(1)) = (V^((2)/(3)))/(((V)/(8))^((2)/(3))) = 8^((2)/(3)) = 4`
So, temperature of gas increases by 4 times.
As mean KE of gas molecule is proportional to its temperature, then KE also increases by 4 times.