The work of 142kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3Jmol^-1K^-1)`

Options

A. triatomic
B. monoatomic
C. diatomic
D. mixture of monoatomic and diatomic

Answer

C

Explanation

Here, `W= -146KJ= -146xx10^(3)J`
`T_(2)-T_(1)= 7^(@)C`
`R= 8.3J "mole"^(-1)K^(-1)`
`=8.3xx10^(3)"J kilo mole"^(-1)K^(-1)`
As `W=(R(T_(2)-T_(1)))/(1-gamma)`
`:. -146xx10^(3)= (8.3xx10^(3)xx7)/(1-gamma)`
`gamma-1=(8.3xx10^(3)xx7)/(146xx10^(3))= 0.40`
`gamma= 1.40`
`:. The gas must be diatomic.