When 2 moles of a monatomic gas expands adiabatically from a temperature of $80^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$, the work done is W . The work done when 3 moles of a diatomic gas expands adiabatically from $50^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$, is

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Aicharya · Accepted Answer

For adiabatic expansion of monoatomic gas $\mathrm{T}_1=80^{\circ} \mathrm{C}$, $\begin{aligned} & \mathrm{T}_2=50^{\circ} \mathrm{C}, \mathrm{n}=2 \text { moles, } \mathrm{f}=3 \\ & \mathrm{~W}=-\Delta \mathrm{u}=-\frac{\mathrm{f}}{2} \mathrm{nR} \Delta \mathrm{~T}=\frac{3}{2} \times 2 \mathrm{R}(80-50) \\ & =90 \mathrm{R} \end{aligned}$
For adiabatic expansion of diatomic gas, $\begin{aligned} & \mathrm{T}_1=50^{\circ} \mathrm{C}, \mathrm{~T}_2=20^{\circ} \mathrm{C}, \mathrm{n}=3 \text { moles, } \mathrm{f}=5 \\ & \therefore \mathrm{~W}^{\prime}=-\frac{\mathrm{f}}{2} \mathrm{nR} \Delta \mathrm{~T}=\frac{5}{2} \times 3 \mathrm{R} \times(50-20) \\ & =\frac{5}{2}(90 \mathrm{R})=2.5 \mathrm{~W} \end{aligned}$

When 2 moles of a monatomic gas expands adiabatically from a temperature of $80^{\circ} \mathrm{C}$ to…

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