You visited us 1 times! Enjoying our articles? Unlock Full Access!

Molar and Specific Heat Capacity

The molar spe...

Question

The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_{p}$ and $C_{v}$ respectively. If $γ = \frac{C_{p}}{C_{v}}$ and $R$ is the universal gas constant, then $C_{v}$ is equal to

\frac{(γ - 1)}{R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

γ R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1 + γ}{1 - γ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R}{(γ - 1)}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{R}{(γ - 1)}$
Using,
$C_{p} - C_{v} = R$
$\Rightarrow C_{v} (\frac{C_{p}}{C_{v}} - 1) = R$
$(γ - 1) = \frac{R}{C_{v}} (∵ \frac{C_{p}}{C_{v}} = γ)$
or $C_{v} = \frac{R}{(γ - 1)}$

Suggest Corrections

Similar questions

C_{p}

C_{v}

γ =

\frac{C_{p}}{C_{v}}

C_{v}

C_{p}

C_{v}

C_{p}

C_{v}

^{'} γ^{'}

C_{v}

C_{p} and C_{v}

γ = \frac{C_{p}}{C_{v}} and R

C_{v}

C_{p}

C_{v}

γ = \frac{C_{p}}{C_{v}}

C_{v}

\frac{C_{p}}{C_{v}} = γ

(C_{v})

Join BYJU'S Learning Program