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 Boyles law, for a given mass of gas, at constant temperature the volume of a gas is inversely proportional to its pressure. \[V\propto \frac{1}{P}\] \[=\frac{dV}{V}\] or PV = constant On differentiating the equation \[P\,(PV)=d\] (constant) \[PdV+VdP=0\] \[VdP=-PdV\] \[-\frac{dV}{V}=\frac{dP}{P}\] \[=-\frac{dV}{V}\]
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