A) \[64p\]
B) \[32p\]
C) \[\frac{p}{64}\]
D) \[16p\]
Correct Answer: C
Solution :
[c] Gas expands isothermally \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[PV={{P}_{2}}(2V)\] \[P=\frac{P}{2}\] It then expands adiabatically \[{{P}_{1}}V_{1}^{\gamma }={{P}_{2}}{{V}_{2}}\] \[\frac{P}{2}{{(2V)}^{\gamma }}={{P}_{2}}{{(16V)}^{\gamma }}\] \[{{P}_{2}}=\frac{P}{2}{{\left[ \frac{2V}{16V} \right]}^{\gamma }}\] \[\gamma =\frac{5}{2}\] [monoatomic] \[{{P}_{2}}=\frac{P}{2}{{\left[ \frac{1}{8} \right]}^{{}^{5}/{}_{3}}}\] \[=\frac{P}{2}{{\left[ \frac{1}{8} \right]}^{3\times {}^{5}/{}_{3}}}\Rightarrow \frac{P}{64}\]| Note | |||
| CP | CV | \[\gamma \] | |
| \[1{}^\circ \] | \[\frac{5R}{2}\] | \[\frac{3R}{2}\] | \[\frac{5}{3}\] |
| \[2{}^\circ \] | \[\frac{7R}{2}\] | \[\frac{5R}{2}\] | \[\frac{7}{5}\] |
| \[3{}^\circ \] | \[\frac{9R}{2}\] | \[\frac{7R}{2}\] | \[\frac{9}{7}\] |
You need to login to perform this action.
You will be redirected in
3 sec
