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Question: Answered & Verified by Expert
If $\mathrm{C}_{\mathrm{p}}$ and $\mathrm{C}_{\mathrm{V}}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight $M$
where $R$ is the molar gas constant.
PhysicsKinetic Theory of GasesNEETNEET 2010 (Mains)
Options:
  • A $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}^2}$
  • B $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R}$
  • C $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}}$
  • D $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{MR}$
Solution:
1441 Upvotes Verified Answer
The correct answer is: $\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}}$
$$
\begin{array}{r}
\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\mathrm{R} \\
\therefore \quad \mathrm{Mc}_{\mathrm{p}}-\mathrm{MC}_{\mathrm{V}}=\mathrm{R} \\
\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{V}}=\frac{\mathrm{R}}{\mathrm{M}}
\end{array}
$$

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