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 $7℃.$ The gas is $\left( R = 8.3\ J\ mol^{- 1}K^{- 1} \right)$
A) Diatomic
B) Triatomic
C) A mixture of monoatomic and diatomic
D) Monoatomic
Explanation
(a) For adiabatic process,
dQ=0
So, $dU = - \Delta W$
$\Rightarrow \ \ \ \ \ nC_{V}dT = + 146 \times 10^{3}J$
$\Rightarrow \ \ \ \ \frac{nfR}{2} \times 7 = 146 \times 10^{3}$
$\lbrack f \rightarrow Degree\ of\ freedom\rbrack$
$\Rightarrow \ \ \frac{10^{3} \times f \times 8.3 \times 7}{2} = 146 \times 10^{3}$
$f\ = 5.02 \approx 5$
So, it is a diatomic gas.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.