Q: 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 = 8.3 J mol^{-1} K^{-1})

(a) Triatomic

(b) A mixture of monoatomic and diatomic

(c) Monoatomic

(d) Diatomic

Sol: (d) n = 10^{3}, W = – 146 × 10^{3} J,

∆T = T_{2} – T_{1} = 7°C

$\large W = \frac{n R (T_2 -T_1)}{1 -\gamma} $

$\large -146 \times 10^3 = \frac{10^3 \times 8.3 \times 7 }{1 -\gamma} $

On solving , we get

γ = 1.4 (diatomic gas )