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

Answer:

(a) V = $\frac{8.3}{P}$

(b) (i) the value of $\frac{dV}{dP}$ when P = 50kPa is - 0.00332 $\frac{m^{3} }{kPa}$

(ii) the meaning of the derivative $\frac{dV}{dP}$ is the rate of change of volume with pressure.

(iii) and the units are $\frac{m^{3} }{kPa}$

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 = $\frac{8.3}{P}$        -------------------(ii)

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

V = $\frac{8.3}{P}$

V = 8.3P⁻¹

$\frac{dV}{dP}$ = -8.3P⁻²

$\frac{dV}{dP}$ = $\frac{-8.3}{P^{2} }$

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

$\frac{dV}{dP}$ = $\frac{-8.3}{50^{2} }$   [$\frac{kPam^{3} }{(kPa)^{2} }$]               -----  Write and evaluate the units alongside

$\frac{dV}{dP}$ = $\frac{-8.3}{2500 }$    [$\frac{kPam^{3} }{k^{2} Pa^{2} }$]

$\frac{dV}{dP}$ = - 0.00332 [$\frac{m^{3} }{kPa}$]

Therefore,

(i) the value of $\frac{dV}{dP}$ when P = 50kPa is - 0.00332 $\frac{m^{3} }{kPa}$

(ii) the meaning of the derivative $\frac{dV}{dP}$ is the rate of change of volume with pressure.

(iii) and the units are $\frac{m^{3} }{kPa}$

Answer 2

Answer:

a)$V=\dfrac{5.3}{P}$

b)$ML^{-4}T^{-2}$.

Explanation:

Given that

Boyle's law

P V = Constant ,at constant temperature

a)

Given that

$P_1=50KPa$

$V_1=0.106m^3$

We know that for PV=C

$P_1V_1=P_2V_2=PV$

Now by putting the values

PV= 50 x 0.106

$V=\dfrac{5.3}{P}$

Where P is in KPa and V is in $m^3$

b)

PV= C

Take ln both sides

So $\ln(PV)=\ln C$

lnP + lnV =lnC               ( C is constant)

By differentiating

$\dfrac{dP}{P}+\dfrac{dV}{V}=0$

So

$\dfrac{dP}{dV}=-\dfrac{P}{V}$

When P= 50 KPa

$\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}$

It indicates the slope of PV=C curve.

It unit is $\dfrac{Pa}{m^3}$.

Or we can say that $ML^{-4}T^{-2}$.