Question

Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P of the gas is inversely proportional to the volume V of the gas. (a) Suppose that the pressure of a sample of air that occupies 0.106 m3 at 25°C is 50 kPa. Write V as a function of P. (b) Calculate dVydP when P − 50 kPa. What is the meaning of the derivative? What are its units?

Answer 1

Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature. To express volume as a function of pressure, use the formula V = k/P. The derivative of volume with respect to pressure indicates the rate at which volume changes as pressure changes, measured in cubic meters per kilopascal (m3/kPa).

Step-by-step explanation:

Boyle’s Law states that for a given amount of gas at a constant temperature, the volume V is inversely proportional to the pressure P. This relationship can be mathematically represented as PV = k, where k is a constant.

(a) Given a sample of air at 25°C with a volume of 0.106 m3 and a pressure of 50 kPa, we can write the volume as a function of pressure by rearranging the formula to V = k/P. Using the initial conditions, we find that k = P × V = 50 kPa × 0.106 m3 = 5.3 kPa · m3, so V(P) = 5.3 kPa · m3 / P.

(b) To calculate dV/dP when P = 50 kPa, we differentiate the function V(P) with respect to P, resulting in dV/dP = -5.3 kPa · m3 / P2. At P = 50 kPa, this becomes dV/dP = -5.3 / (502) = -0.00212 m3/kPa. The derivative represents the rate of change of volume with respect to pressure, which in this case means that for each increase of 1 kPa in pressure, the volume decreases by 0.00212 m3. The units of the derivative are m3/kPa.

Answer 2

(a) V =
`(8.3)/(P)`

(b) (i) the value of
`(dV)/(dP)` when P = 50kPa is - 0.00332
`(m^(3) )/(kPa)`

(ii) the meaning of the derivative
`(dV)/(dP)` is the rate of change of volume with pressure.

(iii) and the units are
`(m^(3) )/(kPa)`

Step-by-step explanation:

Boyle's law states that at constant temperature;

P ∝ 1 / V

=> P = k / V

=> PV = k -------------------------(i)

Where;

P = pressure

V = volume

k = constant of proportionality

According to the question;

When;

V = 0.106m³, P = 50kPa

Substitute these values into equation (i) as follows;

50 x 0.106 = k

Solve for k;

k = 5.3 kPa m³

(a) To write V as a function of P, substitute the value of k into equation (i) as follows;

PV = k

PV = 8.3

Make V subject of the formula in the above equation as follows;

V = 8.3/P

=> V =
`(8.3)/(P)` -------------------(ii)

(b) Find the derivative of equation (ii) with respect to V to get dV/dP as follows;

V =
`(8.3)/(P)`

V = 8.3P⁻¹

`(dV)/(dP)` = -8.3P⁻²

`(dV)/(dP)` =
`(-8.3)/(P^(2) )`

Now substitute P = 50kPa into the equation as follows;

`(dV)/(dP)` =
`(-8.3)/(50^(2) )` [
`(kPam^(3) )/((kPa)^(2) )`] ----- Write and evaluate the units alongside

`(dV)/(dP)` =
`(-8.3)/(2500 )` [
`(kPam^(3) )/(k^(2) Pa^(2) )`]

`(dV)/(dP)` = - 0.00332 [
`(m^(3) )/(kPa)`]

Therefore,

(i) the value of
`(dV)/(dP)` when P = 50kPa is - 0.00332
`(m^(3) )/(kPa)`

(ii) the meaning of the derivative
`(dV)/(dP)` is the rate of change of volume with pressure.

(iii) and the units are
`(m^(3) )/(kPa)`