A) \[4pV+3TA=0\]
B) \[3pV-4TA=0\]
C) \[4pV-3TA=0\]
D) \[3pV+4TA=0\]
Correct Answer: D
Solution :
\[{{n}_{1}}+{{n}_{1}}={{n}_{2}}\] \[\frac{{{P}_{1}}{{V}_{1}}}{RT}+\frac{{{P}_{1}}{{V}_{1}}}{RT}=\frac{{{P}_{2}}{{V}_{2}}}{RT}\] \[2{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[2\left( P+\frac{4T}{r} \right)\,\frac{4}{3}\,\pi {{r}^{3}}\,=\left( P+\frac{4T}{R} \right)\frac{4}{3}\,\pi {{R}^{3}}\] \[0=P\,\left( \frac{4}{3}\,\pi {{R}^{3}}\,-\frac{4}{3}\,\pi {{r}^{3}}\times 2 \right)\,+\frac{4T}{3}\,\left( \frac{4}{3}\,\pi {{R}^{2}}-4\pi {{\rho }^{2}}\times 2 \right)\]\[0=PV+\frac{4TA}{3}\] \[0=3PV\,+4T\,A\]
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