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Question: Equation of Boyle’s law: A) \(dP/P = - dV/V\) B) \(dP/P = + dV/V\) C) \({d^2}P/P = - dV/V\) ...

Equation of Boyle’s law:
A) dP/P=−dV/VdP/P = - dV/V
B) dP/P=+dV/VdP/P = + dV/V
C) d2P/P=−dV/V{d^2}P/P = - dV/V
D) d2P/P=+dV/V{d^2}P/P = + dV/V

Explanation

Solution

Hint : Boyle’s law is also known as the pressure-volume relationship which states that when the temperature is constant the pressure of a fixed amount of gas varies inversely with its volume.

_Complete step by step solution : _
-We should first start by seeing what Boyle’s law is: Boyle’s law is an experimental gas law that describes the relationship between pressure and volume when the temperature is constant. It states that pressure of a gas tends to increase when the volume of a container decreases.
It can be written as: Pα1/VP\alpha 1/V (Temperature is constant)
P=k(1/V)P = k(1/V)
PV=kPV = k
where k is proportionality constant whose value depends on the temperature of the gas, amount of gas and units of P and V.
So, for a gas at temperature T, occupying volume V1 at pressure P1 and volume V2 at pressure P2, Boyle’s law can be written as:
P1V1=P2V2=k{P_1}{V_1} = {P_2}{V_2} = k
P1/P2=V2/V1{P_1}/{P_2} = {V_2}/{V_1}
-Now, we use the equation: PV=kPV = k, to derive the differentiated form of Boyle’s Law. Differentiate the equation: PV=kPV = k .
By differentiating it we get: dP/P+dV/V=0dP/P + dV/V = 0
So, dP/P=−dV/VdP/P = - dV/V

So, the correct option is, (A) dP/P=−dV/VdP/P = - dV/V

Note : The Boyle’s law formula: P1/P2=V2/V1{P_1}/{P_2} = {V_2}/{V_1}; is valid for only a gas at a constant temperature but varying pressure and temperature and not for 2 different gases.