Question

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?

Answer 1

Answer:

For these types of questions the equation that we must take into account is that:

T = PxV (where T is the temperature, P is the pressure and V is the volume) this equation is described as we consider that this is the value N and R is 1, therefore it is not necessary to explain them now.

Explanation:

The quoted equation refers to Boyle's Law, in this law we can explain that the volume increases if the pressure decreases and if the temperature also increases, if the pressure increases and the volume decreases this means that the gas is compressing assuming that the temperature is constant

Answer 2

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' = $-\frac{C}{P^2}$.

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 = $\frac{C}{P}$

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

$\frac{dV}{dt} = -\frac{C}{P^2} \frac{dP}{dt}$ which gives

$\frac{dV}{dt} \times \frac{dt}{dP}= \frac{dV}{dP} = -\frac{C}{P^2}$

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.

$\frac{dV}{dP} = -\frac{C}{P^2}$.

V' = $-\frac{C}{P^2}$.