A) \[\frac{dp}{p}=-\frac{dV}{V}\]
B) \[\frac{dp}{p}=+\frac{dV}{V}\]
C) \[\frac{{{d}^{2}}p}{p}=-\frac{dV}{dT}\]
D) \[\frac{{{d}^{2}}p}{p}=+\frac{{{d}^{2}}V}{dT}\]
Correct Answer: A
Solution :
According to Boyle's law, "for a given mass of a gas, at constant temperature the volume of a gas is inversely proportional to its pressure." \[V\propto \frac{1}{P}\] or PV = constant on differentiating the equation \[d(PV)=d\] (constant) \[\Rightarrow \] \[pdV+VdP=0\] \[=VdP=-PdV\] \[=\frac{dp}{p}=-\frac{dV}{V}\]
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