Work done on heating one mole of monoatomic g | vedclass

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(B) The work done in an adiabatic process is given by the formula: $\Delta W = \frac{\mu R \Delta T}{\gamma - 1}$.For $1$ mole of monoatomic gas $(\mu

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$10$

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(B) The work done in an adiabatic process is given by the formula: $\Delta W = \frac{\mu R \Delta T}{\gamma - 1}$.For $1$ mole of monoatomic gas $(\mu = 1)$: $\Delta T = 20^{\circ} C$, $\gamma = 5/3$. Therefore, $W = \frac{1 	imes R 	imes 20}{(5/3 - 1)} = \frac{20R}{2/3} = 30R$ ... $(i)$.For $6$ moles of rigid diatomic gas $(\mu = 6)$: $\Delta T = 20^{\circ} C$, $\gamma = 7/5$. The work done $W'$ is given by: $W' = \frac{6 	imes R 	imes 20}{(7/5 - 1)} = \frac{120R}{2/5} = \frac{120R 	imes 5}{2} = 300R$.Comparing $W'$ with $W$: $W' = 300R = 10 	imes (30R) = 10W$.

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 in temperature is: (in $W$)

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