If the pressure P applied to a gas is increased while the gas is held at a constant temperature, then the volume V of the gas will decrease. 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. Which of the following is a differential equation that could describe this relationship?

Answers

Answer 1

Answer:

The differential relationship has been $\rm \bold{(dV)/(dP)\;=\;(C)/(P^2)}$ .

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 $*$ Volume = Constant

PV = C

V = $\rm (C)/(P)$

Differentiating the equation:

$\rm (dV)/(dt)\;=\;(C)/(P^2)\;(dP)/(dt)$

$\rm (dV)/(dt)\;*\;(dt)/(dP)\;=\;(C)/(P^2)$

$\rm (dV)/(dP)\;=\;(C)/(P^2)$

The differential relationship has been $\rm \bold{(dV)/(dP)\;=\;(C)/(P^2)}$ .

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Answer 2

Answer:

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' = $-(C)/(P^2)$ .

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 = $(C)/(P)$

Therefore, the rate of change of volume of a gas is given as

$(dV)/(dt) = -(C)/(P^2) (dP)/(dt)$ which gives

$(dV)/(dt) * (dt)/(dP)= (dV)/(dP) = -(C)/(P^2)$

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.

$(dV)/(dP) = -(C)/(P^2)$ .

V' = $-(C)/(P^2)$ .

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Answers

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