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)

The work of 146 kJ is performed in order to compress one kilo mole of

1.	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)
Answer» Triatomic Monoatomic Diatomic Mixture of monoatomic & diatomic Solution :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.