When 40 J of heat is absorbed by a monatomic gas, the increase in the internal energy of the gas is

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Aicharya · Accepted Answer

For mono atomic gas, $\mathrm{f}=3$ $	herefore V=1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}$ $\begin{aligned} & 	ext { using, } \mathrm{Q}=\mathrm{W}+\Delta \mathrm{U}=\mathrm{nR} \Delta \mathrm{T}+\mathrm{nc}_{\mathrm{v}} \Delta \mathrm{T}=\mathrm{nc}_{\mathrm{p}} \Delta \mathrm{T} \ & 	herefore \frac{\Delta \mathrm{U}}{\mathrm{Q}}=\frac{\mathrm{nc}_{\mathrm{v}} \Delta \mathrm{T}}{\mathrm{nc}_{\mathrm{p}} \Delta \mathrm{T}}=\frac{1}{V}\end{aligned}$ $\Rightarrow \Delta U=\frac{	heta}{8}=\frac{40}{5 / 3}=24 \mathrm{~J}$

When 40 J of heat is absorbed by a monatomic gas, the increase in the internal energy of the gas is

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