The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3ml^-1Jmol^-1K^-1)`

Options

A. Triatomic
B. Monoatomic
C. Diatomic
D. Mixture of monoatomic & diatomic

Answer

C

Explanation

Here `W = -146 KJ = –146 × 10^3J`
`T_2 – T_1 = 7^@C, R = 8.31 "mole"^(-1) K^(-1)`
` = 8.3 xx 10^(3) "J kilomole"^(-1) (K^-1)`
As `w= (R(T_2 - T_1))/(1-gamma)`
`-146.xx10^(3) = (8.3xx10^3xx7)/(1-gamma)`
`gamma-1=(8.3xx10^3xx7)/(146xx10^3)=0.40`
`gamma=1.40` The gas must be diatomic.