A) 64 p
B) 32 p
C) \[\frac{p}{64}\]
D) 16 p
Correct Answer: C
Solution :
For isothermal expansion |
\[pV=p'\times 2V\] \[[\therefore V'=2V]\] |
\[p'=\frac{p}{2}\] |
For adiabatic expansion, |
\[p{{V}^{\gamma }}=\] constant |
\[p'{{V}^{'\gamma }}=p''V'{{'}^{\gamma }}\] |
\[\frac{p}{2}{{[2V]}^{5/3}}=p''{{[16V]}^{5/3}}\] |
\[p''=\frac{p}{2}{{\left[ \frac{2V}{16V} \right]}^{5/3}}=\frac{p}{2}{{\left[ \frac{1}{8} \right]}^{5/3}}\] |
\[=\frac{p}{2}[0.03125]\] |
\[=0.0156p\] |
\[=p/64\] |
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