11.Thermodynamics
hard

Consider that an ideal gas ($n$ moles) is expanding in a process given by $P = f (V)$, which passes through a point $(V_0, \,p_0)$. Show that the gas is absorbing heat at $(p_0,\, V_0)$ if the slope of the curve $P = f (V)$ is larger than the slope of the adiabatic passing through $(p_0,\, V_0)$.

Option A
Option B
Option C
Option D

Solution

Slope of graph $\mathrm{P}=f(\mathrm{~V})$ at point $\left(\mathrm{V}_{0}, p_{0}\right)=f\left(\mathrm{~V}_{0}\right)$. For adiabatic process $\mathrm{PV}^{\gamma}=\mathrm{K}$ (Constant)

$\therefore$ At point $\left(\mathrm{V}_{0}, \mathrm{P}_{0}\right) \mathrm{P}_{0} \mathrm{~V}_{0}^{\gamma}=\mathrm{k}$

$\therefore \mathrm{P}_{0}=\frac{k}{\mathrm{~V}_{0}^{\gamma}}$

$\therefore d \mathrm{P}_{0}=k(-\gamma) \mathrm{V}_{0}^{-\gamma-1} \cdot d \mathrm{~V}_{0}$

$\therefore \frac{d \mathrm{P}_{0}}{d \mathrm{~V}_{0}}=-\gamma \frac{k}{\mathrm{~V}_{0}^{\gamma}} \times \frac{1}{\mathrm{~V}_{0}}$

$\therefore$ Slope $=-\gamma \frac{\mathrm{P}_{0}}{\mathrm{~V}_{0}}$

Now $d \mathrm{Q}=d \mathrm{U}+d \mathrm{~W} \quad\left[\because \frac{k}{\mathrm{~V}_{0}^{\gamma}}=\mathrm{P}_{0}\right]$ $…..(2)$

and $\mathrm{PV}=n \mathrm{RT}$

$\therefore \mathrm{T}=\frac{\mathrm{PV}}{n \mathrm{R}}=\frac{f(\mathrm{~V}) \mathrm{V}}{n \mathrm{R}}$

$\therefore d \mathrm{~T}=\frac{1}{n \mathrm{R}}\left[f(\mathrm{~V}) d \mathrm{~V}+\mathrm{V} f^{\prime}(\mathrm{V}) d \mathrm{~V}\right]$

$\quad=\frac{1}{n \mathrm{R}}\left[f(\mathrm{~V})+\mathrm{V} f^{\prime}(\mathrm{V})\right] d \mathrm{~V} \ldots . .(3)$

$\Rightarrow \mathrm{Equ} .(2) d \mathrm{Q}=n \mathrm{C}_{\mathrm{V}} \times \frac{1}{n \mathrm{R}}\left[f(\mathrm{~V})+\mathrm{V} f^{\prime}(\mathrm{V})\right] d \mathrm{~V}$

$\therefore\left[\frac{d \mathrm{Q}}{d \mathrm{~V}}\right]_{\mathrm{V}}=\mathrm{V}_{0}$

Now, $\mathrm{C}_{\mathrm{P}}-\frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{R}}\left[f\left(\mathrm{~V}_{\mathrm{V}}\right)+\mathrm{R}_{0} f^{\prime}\left(\mathrm{V}_{0}\right)\right]+f\left(\mathrm{~V}_{0}\right)$

$\therefore \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}$

$\quad \gamma-1=\frac{\mathrm{R}}{\mathrm{C}_{\mathrm{V}}}$

$\therefore \frac{\mathrm{C}_{\mathrm{V}}}{\mathrm{R}}=\frac{1}{\gamma-1}$ $….(5)$

Standard 11
Physics

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