A) 64p
B) 32p
C) \[\frac{p}{64}\]
D) 16p
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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