One mole of an ideal gas passes through a process where

Question:

One mole of an ideal gas passes through a process where

pressure and volume obey the relation $P=P_{0}\left[1-\frac{1}{2}\left(\frac{V_{0}}{V}\right)^{2}\right]$.

Here $P_{o}$ and $V_{o}$ are constants. Calculate the charge in the temperature of the gas if its volume changes from $\mathrm{V}_{\mathrm{o}}$ to $2 \mathrm{~V}_{\mathrm{o}}$.

 

  1. (1) $\frac{1}{2} \frac{\mathrm{Po} \mathrm{V}_{o}}{\mathrm{R}}$

  2. (2) $\frac{5}{4} \frac{\mathrm{P}_{o} \mathrm{~V}_{o}}{\mathrm{R}}$

  3. (3) $\frac{3}{4} \frac{\mathrm{P}_{\mathrm{o}} \mathrm{V}_{\mathrm{o}}}{\mathrm{R}}$

  4. (4) $\frac{1}{4} \frac{\mathrm{P}_{o} \mathrm{~V}_{o}}{\mathrm{R}}$


Correct Option: , 2

Solution:

(2) We have given,

$P=P_{0}\left[1-\frac{1}{2}\left(\frac{V_{0}}{V}\right)^{2}\right]$

When $\mathrm{V}_{1}=\mathrm{V}_{0}$

$\Rightarrow P_{1}=P_{0}\left[1-\frac{1}{2}\right]=\frac{P_{0}}{2}$

When $\mathrm{V}_{2}=2 \mathrm{~V}_{0}$

$\Rightarrow P_{2}=P_{0}\left[1-\frac{1}{2}\left(\frac{1}{4}\right)\right]=\left(\frac{7 P_{0}}{8}\right)$

$\Delta T=T_{2}-T_{1}=\left|\frac{P_{1} V_{1}}{n R}-\frac{P_{2} V_{2}}{n R}\right|$                             $\left[\because T=\frac{P V}{n R}\right]$

$\Delta T=\left|\left(\frac{1}{n R}\right)\left(P_{1} V_{1}-P_{2} V_{2}\right)\right|=\left(\frac{1}{n R}\right)\left|\left(\frac{P_{0} V_{0}}{2}-\frac{7 P_{0} V_{0}}{4}\right)\right|$

$=\frac{5 P_{0} V_{0}}{4 n R}=\frac{5 P_{0} V_{0}}{4 R} \quad(\because n=1)$

 

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