Question

The work of $146kJ$ is performed in order to compress one kile mole of gas adiabatically and in this process the temperature of the gas increases by $7^{o}C$. The gas is
$(R=8.3J{mol}^{-1}{K}^{-1})$

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

Answer 1

Answer: (B)

diatomic

$\text{Given:-}$ Work done $=146J$

$R=8.3Jmo{l}^{-1}{K}^{-1}$

$\text{Solution:-}$

For adiabatic process, no exchange of heat

$\therefore$ $dQ=0$

So, $dU=-\Delta W$

$\Rightarrow n{C}_{V}dT=+146\times {10}^{3}J$

$\Rightarrow \frac{nfR}{2}\times =146\times {10}^3$

[where, f is degree of freedom]

$\Rightarrow \frac{{10}^3\times f\times 8.3\times 7}{2}=146\times {10}^3$

$\therefore f=5.02\approx 5$

So, it is a diatomic gas

$\text{Hence the correct option is B}$