A) \[\frac{dp}{p}=-\frac{dV}{V}\]
B) \[\frac{dp}{p}=+\frac{dV}{V}\]
C) \[\frac{{{d}^{2}}p}{p}=-\frac{dV}{V}\]
D) \[\frac{{{d}^{2}}p}{p}=+\frac{{{d}^{2}}V}{dT}\]
Correct Answer: A
Solution :
Boyles law pV = constant On differentiating the equation, \[d(pV)=d(c)\Rightarrow pdV+Vdp=0\] \[\Rightarrow \] \[Vdp=-pdV\Rightarrow \frac{dp}{p}=\frac{-dV}{V}\]
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