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An ideal gas undergoes a quasi intatic, reversible process in which ito molar heat capacity $C$ remain constant. If during this process the relation of pressure $P$ and volume $V$ is given ly $p^{V^N}=$ constant, then $N$ in given dy $($ here $C_p$ and $C_v$ are molar specific heat et constant pressure ad constant volume, respectively $)$
Option: 1
$N=\frac{c_P}{c_V}$

Option: 2
$N=\frac{c-c_p}{c-c_v}$

Option: 3
$N=\frac{C_p{-c}}{c-C_v}$

Option: 4
$N=\frac{C-C_v}{C-C_p}$

Answers (1)

best_answer

Here, $\left.P V^N=K \text { (Constant }\right)$

For $1mol$ of ideal gas, $P^V=R T$

$\text { Dividing (1) } by\text { (2) We get, } V^N-1_T=\frac{K}{R}$

$\therefore\left(\frac{d V}{d T}\right)=\frac{V}{(N-1)^T}=\frac{V}{(1-N)^T}$

According to the first law of thermodynamics,

$\begin{aligned} d Q & =C_V d T+p^{d V} \\ \therefore \frac{d Q}{d T} & =C_V+p\left(\frac{dV}{d T}\right)=C_V+\frac{p^V}{(1-N) T}=C_V+\frac{R}{1-N} \end{aligned}$

Hence, thermal capacity $C=C_v+\frac{R}{1-N} \text { or, } 1-N=\frac{{R}}{C} C_v$

$\begin{aligned} &\text { or, } N=1-\frac{R}{c-c_v}=\frac{c-(c_v+R)}{c-c_v}=\frac{c-c_p}{C-c_v}\\ &\left[\because C_p-c_V=R\right] \end{aligned}$

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Anam Khan

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