A monoatomic gas is heated at constant pressure. The percentage of total heat used for doing external work is

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Aicharya · Accepted Answer

For constant pressure, Using first law of thermodynamics, $\begin{aligned} & \mathrm{W}=\mathrm{Q}-\Delta \mathrm{U} \ & \mathrm{~W}=\mathrm{mC}_{\mathrm{p}} \Delta \mathrm{~T}-\mathrm{mC}_{\mathrm{v}} \Delta \mathrm{~T} \end{aligned}$ For monoatomic gas $\gamma=\frac{C_p}{C_v}=\frac{5}{3}$ $\begin{aligned} & \quad \Rightarrow C_v=\frac{3}{5} C_p \ & 	herefore \quad W=m C_p \Delta T-m\left(\frac{3}{5} C_pight) \Delta T \ & W=\frac{2}{5} m_p \Delta T=\frac{2}{5} Q \Rightarrow 40 \% 	ext { of heat }\end{aligned}$ ^

A monoatomic gas is heated at constant pressure. The percentage of total heat used for doing external work is

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