The ratio $\frac{C_{p}}{C_{c}} = γ$ for gas. Its molecular weight is 'M'. Its specific heat capacity at constant pressure is :

\frac{R}{γ - 1}

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\frac{γ R}{γ - 1}

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\frac{γ R}{M (γ - 1)}

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\frac{γ R M}{(γ - 1)}

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Solution

The correct option is C $\frac{γ R}{M (γ - 1)}$
Since $C_{P} - C_{V} = \frac{R}{M}$ and $\frac{C_{P}}{C_{V}} = γ$ ,

$C_{V} = \frac{R}{M (γ - 1)}$

and $C_{P} = \frac{γ R}{M (γ - 1)}$

Now, at constant pressure, $Q = Δ H = n C_{P} Δ T = \frac{γ R}{M (γ - 1)}$

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Similar questions

For an ideal gas $(\frac{C_{P}}{C_{v}} = γ)$ ; of molar mass M, its specific heat capacity at constant volume is:

(γ = \frac{C_{p}}{C_{V}})

For an ideal gas $(\frac{C_{P}}{C_{v}} = γ)$ ; of molar mass M, its specific heat capacity at constant volume is:

\frac{C p, m}{C v, m} = γ

\frac{C_{p}}{C_{v}} = γ

(C_{v})

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