Respuesta :
Answer:
A differential equation that could describe the relationship of the rate of change of the volume of gas with respect to the pressure is;
V' = [tex]-\frac{C}{P^2}[/tex].
Explanation:
Boyle's law states that at constant temperature, the pressure of a given mass of gas is inversely proportional to its volume.
That is;
P₁×V₁ = P₂×V₂ or
P×V = Constant, C
That is V = [tex]\frac{C}{P}[/tex]
Therefore, the rate of change of volume of a gas is given as
[tex]\frac{dV}{dt} = -\frac{C}{P^2} \frac{dP}{dt}[/tex] which gives
[tex]\frac{dV}{dt} \times \frac{dt}{dP}= \frac{dV}{dP} = -\frac{C}{P^2}[/tex]
That is the rate of change of the volume of gas with respect to the pressure is proportional to the reciprocal of the square of the pressure.
[tex]\frac{dV}{dP} = -\frac{C}{P^2}[/tex].
V' = [tex]-\frac{C}{P^2}[/tex].
The differential relationship has been [tex]\rm \bold{\dfrac{dV}{dP}\;=\;\dfrac{C}{P^2}}[/tex].
The gas has been termed to be the ideal gas. For an ideal gas at a constant temperature, the relationship of the change in pressure and volume can be given as constant. The relationship has been given with the application of Boyle's law.
The product of the pressure and volume has been a constant quantity for a reaction.
Pressure [tex]\times[/tex] Volume = Constant
PV = C
V = [tex]\rm \dfrac{C}{P}[/tex]
Differentiating the equation:
[tex]\rm \dfrac{dV}{dt}\;=\;\dfrac{C}{P^2}\;\dfrac{dP}{dt}[/tex]
[tex]\rm \dfrac{dV}{dt}\;\times\;\dfrac{dt}{dP}\;=\;\dfrac{C}{P^2}[/tex]
[tex]\rm \dfrac{dV}{dP}\;=\;\dfrac{C}{P^2}[/tex]
The differential relationship has been [tex]\rm \bold{\dfrac{dV}{dP}\;=\;\dfrac{C}{P^2}}[/tex].
For more information about pressure at a constant temperature, refer to the link:
https://brainly.com/question/12152879?
