A) \[\frac{5{{p}_{0}}}{6\alpha }\]
B) \[\frac{{{p}_{0}}}{2\alpha }\]
C) \[\frac{{{p}_{0}}}{4\alpha }\]
D) \[\frac{5{{p}_{0}}}{8\alpha }\]
Correct Answer: D
Solution :
[d] \[ds=n{{C}_{v}}dT+PdV=0\] \[nR\frac{dT}{dV}+\left( {{p}_{0}}-\alpha V \right)=0\] \[pV=nRT\] \[{{p}_{0}}V-\alpha {{V}^{2}}=nRT\] \[{{p}_{0}}-2\alpha V=nR\frac{dT}{dV}\] \[-\left( {{p}_{0}}-\alpha V \right)\left( \gamma -1 \right)={{p}_{0}}-2\alpha V\] \[-{{p}_{0}}(\gamma -1)+\alpha (\gamma -1)\,V={{p}_{0}}-2\alpha V\] \[{{p}_{0}}V=\alpha V(\gamma +1)\] \[V=\frac{{{p}_{0}}\gamma }{\alpha \left( \gamma +1 \right)}\] \[V=\frac{{{p}_{0}}\times \frac{5}{3}}{\alpha \left( \frac{5}{3}+1 \right)}=\frac{5{{p}_{0}}}{8\alpha }\]You need to login to perform this action.
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