A) \[\frac{1}{2}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
B) \[\frac{3}{4}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
C) \[\frac{5}{4}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
D) \[\frac{1}{4}\frac{{{P}_{0}}{{V}_{0}}}{R}\]
Correct Answer: C
Solution :
| [c] \[P={{P}_{0}}\left[ 1-\frac{1}{2}{{\left( \frac{{{V}_{0}}}{V} \right)}^{2}} \right]\] |
| Pressure at \[{{V}_{0}}={{P}_{0}}\left( 1-\frac{1}{2} \right)=\frac{{{P}_{0}}}{2}\] |
| Pressure at\[2{{V}_{0}}={{P}_{0}}\left( 1-\frac{1}{2}\times \frac{1}{4} \right)=\frac{7}{8}{{P}_{0}}\] |
| Temperature at \[{{V}_{0}}=1-\frac{\frac{{{P}_{0}}}{2}{{V}_{0}}}{nR}=\frac{{{P}_{0}}{{V}_{0}}}{2nR}\] |
| Temperature at \[2{{V}_{0}}=\frac{\left( \frac{7}{8}{{P}_{0}} \right)(2{{V}_{0}})}{nR}=\frac{\frac{7}{4}{{P}_{0}}{{V}_{0}}}{nR}\] |
| Change in temperature\[=\left( \frac{7}{4}-\frac{1}{2} \right)\frac{{{P}_{0}}{{V}_{0}}}{nR}\] |
| \[=\frac{5}{4}\frac{{{P}_{0}}{{V}_{0}}}{nR}=\frac{5{{P}_{0}}{{V}_{0}}}{4R}\] |
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