Equation of Boyle's law is:

Question

Equation of Boyle's law is: (A)  (dp/p)=-(dV/V)  (B)  (dp/p)=+(dV/V)  (C)  (d2p/p)=-(dV/dT)  (D)  (d2p/p)=+(d2V/dT)

Tardigrade Admin · Accepted Answer

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 (1/P) or PV = constant on differentiating the equation d(PV)=d (constant) ⇒ pdV+VdP=0 =VdP=-PdV =(dp/p)=-(dV/V)

Q. Equation of Boyle's law is:

$\frac{d p}{p} = - \frac{d V}{V}$

$\frac{d p}{p} = + \frac{d V}{V}$

$\frac{d ^{2} p}{p} = - \frac{d V}{d T}$

$\frac{d ^{2} p}{p} = + \frac{d ^{2} V}{d T}$

Q. Equation of Boyle's law is:

pdp​=−VdV​

pdp​=+VdV​

pd2p​=−dTdV​

pd2p​=+dTd2V​

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∝P1​ or PV = constant on differentiating the equation d(PV)=d (constant) ⇒ pdV+VdP=0 =VdP=−PdV =pdp​=−VdV​

$\frac{d p}{p} = - \frac{d V}{V}$

$\frac{d p}{p} = + \frac{d V}{V}$

$\frac{d ^{2} p}{p} = - \frac{d V}{d T}$

$\frac{d ^{2} p}{p} = + \frac{d ^{2} V}{d T}$