For an ideal gas the instantaneous change in pressure

Question:

For an ideal gas the instantaneous change in pressure ' $p$ ' with volume ' $v$ ' is given by the

equation $\frac{\mathrm{dp}}{\mathrm{dv}}=-\mathrm{ap}$. If $\mathrm{p}=\mathrm{p}_{0}$ at $\mathrm{v}=0$ is the given

boundary condition, then the maximum temperature one mole of gas can attain is :

(Here $R$ is the gas constant)

  1. $\frac{\mathrm{p}_{0}}{\mathrm{aeR}}$

  2. $\frac{\mathrm{ap}_{0}}{\mathrm{eR}}$

  3. infinity

  4. $0^{\circ} \mathrm{C}$


Correct Option: 1

Solution:

$\int_{p_{0}}^{p} \frac{d p}{p}=-a \int_{0}^{v} d v$

$\ln \left(\frac{\mathrm{p}}{\mathrm{p}_{0}}\right)=-\mathrm{av}$

$\mathrm{p}=\mathrm{p}_{0} \mathrm{e}^{-\mathrm{av}}$

For temperature maximum p-v product should be maximum

$\mathrm{T}=\frac{\mathrm{pv}}{\mathrm{nR}}=\frac{\mathrm{p}_{0} \mathrm{ve}^{-\mathrm{av}}}{\mathrm{R}}$

$\frac{\mathrm{dT}}{\mathrm{dv}}=0 \Rightarrow \frac{\mathrm{p}_{0}}{\mathrm{R}}\left\{\mathrm{e}^{-\mathrm{av}}+\mathrm{ve}^{-\mathrm{av}}(-\mathrm{a})\right\}$

$\frac{\mathrm{p}_{0} \mathrm{e}^{-\mathrm{av}}}{\mathrm{R}}\{1-\mathrm{av}\}=0$

$\mathrm{v}=\frac{1}{\mathrm{a}}, \infty$

$\mathrm{T}=\frac{\mathrm{p}_{0} 1}{\mathrm{Rae}}=\frac{\mathrm{p}_{0}}{\mathrm{Rae}}$

at $v=\infty$

$\mathrm{T}=0$

Option (1)

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now