If ' CP ' and ' CV ' are molar specific heats of an ideal gas at constant pressure and volume respectively. If ' γ ' is the ratio of two specific heats and ' R ' is universal gas constant then ' Cp ' is equal to (a) (Rγ)/(γ-1) (b) γ R (c) (1+γ)/(1-γ) (d) (R)/(γ-1)

Answer: (a). Cp - Cv = R with Cv = (Cp)/(γ). Cp·(γ-1)/(γ) = R ⇒ Cp = (Rγ)/(γ-1) So the answer is (a) (Rγ)/(γ-1).

If ' CP ' and ' CV ' are molar specific heats of an ideal gas at… | MHT CET 2019 (2 May, Shift 1)

Q49MHT CET 2019 · 2 MAY · SHIFT 1Easy

Specific Heats of Gases - Cp, Cv, γ & Mayer's Relation

If ' $C_P$ ' and ' $C_V$ ' are molar specific heats of an ideal gas at constant pressure and volume respectively. If ' $\gamma$ ' is the ratio of two specific heats and ' $R$ ' is universal gas constant then ' $C_p$ ' is equal to

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Answer: (a)

$C_p - C_v = R$ with $C_v = \frac{C_p}{\gamma}$ .
$C_p\cdot\frac{\gamma-1}{\gamma} = R \Rightarrow C_p = \frac{R\gamma}{\gamma-1}$
So the answer is (a) $\frac{R\gamma}{\gamma-1}$ .

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Specific Heats of Gases - Cp, Cv, γ & Mayer's Relation question from MHT CET 2019 (2 May, Shift 1)

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