Question

Work done on heating one mole of monoatomic gas adiabatically through $20^{\circ} C$ is W.Then the work done on heating $6$ moles of rigid diatomic gas through the same change intemperature

• A. 9 W
• B. 10 W
• C. 12 W
• D. 8 W

Answer 1

Answer: (B)

Work done in an adiabatic expansion is
$\Delta W=\frac{\mu R \Delta T}{\gamma-1}$
For $1$ mole of monoatomic gas,
$\Delta T=20^{\circ} C, \gamma=\frac{5}{3}$
$\therefore W=\frac{1 \times R \times 20}{\left(\frac{5}{3}-1\right)}=30 R \ldots \text { (i) }$
For $6$ moles of rigid diatomic gas,
$\Delta T=20^{\circ} C , \gamma=\frac{7}{5} $
$\Delta W'=\frac{6 \times R \times 20}{\left(\frac{7}{5}-1\right)}=\frac{5 \times 6 \times R \times 20}{2} $
$=10 \times 30 R=10 W $ [ from Eq. (i) ]