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,
$\Delta T=20^{\circ }C,\gamma =\frac{5}{3}$
$\therefore W=\frac{1\times R\times 20}{\left(\frac{5}{3}-1\right)}=30R\dots \text{ (i) }$
For $6$ moles of rigid diatomic gas,
$\Delta T=20^{\circ }C,\gamma =\frac{7}{5}$
$\Delta W^{\prime }=\frac{6\times R\times 20}{\left(\frac{7}{5}-1\right)}=\frac{5\times 6\times R\times 20}{2}$
$=10\times 30R=10W$ [ from Eq. (i) ]