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 :
Boyle's law \[-PV\]= constant On differentiating the equation, \[d(PV)=d(C)\] Þ \[PdV+VdP=0\] Þ \[VdP=-PdV\]Þ \[\frac{dP}{P}=-\frac{dV}{V}\].
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