MHT CET · Physics · Thermal Properties of Matter
A monoatomic gas is heated at constant pressure. The percentage of total heat used for doing external work is
- A \(30 \%\)
- B \(40 \%\)
- C \(50 \%\)
- D \(60 \%\)
Answer & Solution
Correct Answer
(B) \(40 \%\)
Step-by-step Solution
Detailed explanation
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 \(\gamma=\frac{C_p}{C_v}=\frac{5}{3}\)
\(\begin{aligned} & \quad \Rightarrow C_v=\frac{3}{5} C_p \\ & \therefore \quad W=m C_p \Delta T-m\left(\frac{3}{5} C_p\right) \Delta T \\ & W=\frac{2}{5} m_p \Delta T=\frac{2}{5} Q \Rightarrow 40 \% \text { of heat }\end{aligned}\)
\(\begin{aligned}
& \mathrm{W}=\mathrm{Q}-\Delta \mathrm{U} \\
& \mathrm{~W}=\mathrm{mC}_{\mathrm{p}} \Delta \mathrm{~T}-\mathrm{mC}_{\mathrm{v}} \Delta \mathrm{~T}
\end{aligned}\)
For monoatomic gas \(\gamma=\frac{C_p}{C_v}=\frac{5}{3}\)
\(\begin{aligned} & \quad \Rightarrow C_v=\frac{3}{5} C_p \\ & \therefore \quad W=m C_p \Delta T-m\left(\frac{3}{5} C_p\right) \Delta T \\ & W=\frac{2}{5} m_p \Delta T=\frac{2}{5} Q \Rightarrow 40 \% \text { of heat }\end{aligned}\)
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