Respuesta :
Answer:
a)[tex]V=\dfrac{5.3}{P}[/tex]
b)[tex]ML^{-4}T^{-2}[/tex].
Explanation:
Given that
Boyle's law
P V = Constant ,at constant temperature
a)
Given that
[tex]P_1=50KPa[/tex]
[tex]V_1=0.106m^3[/tex]
We know that for PV=C
[tex]P_1V_1=P_2V_2=PV[/tex]
Now by putting the values
PV= 50 x 0.106
[tex]V=\dfrac{5.3}{P}[/tex]
Where P is in KPa and V is in [tex]m^3[/tex]
b)
PV= C
Take ln both sides
So [tex]\ln(PV)=\ln C[/tex]
lnP + lnV =lnC ( C is constant)
By differentiating
[tex]\dfrac{dP}{P}+\dfrac{dV}{V}=0[/tex]
So
[tex]\dfrac{dP}{dV}=-\dfrac{P}{V}[/tex]
When P= 50 KPa
[tex]\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}[/tex]
It indicates the slope of PV=C curve.
It unit is [tex]\dfrac{Pa}{m^3}[/tex].
Or we can say that [tex]ML^{-4}T^{-2}[/tex].
Answer:
(a) V = [tex]\frac{8.3}{P}[/tex]
(b) (i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]
(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.
(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]
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 = [tex]\frac{8.3}{P}[/tex] -------------------(ii)
(b) Find the derivative of equation (ii) with respect to V to get dV/dP as follows;
V = [tex]\frac{8.3}{P}[/tex]
V = 8.3P⁻¹
[tex]\frac{dV}{dP}[/tex] = -8.3P⁻²
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{P^{2} }[/tex]
Now substitute P = 50kPa into the equation as follows;
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{50^{2} }[/tex] [[tex]\frac{kPam^{3} }{(kPa)^{2} }[/tex]] ----- Write and evaluate the units alongside
[tex]\frac{dV}{dP}[/tex] = [tex]\frac{-8.3}{2500 }[/tex] [[tex]\frac{kPam^{3} }{k^{2} Pa^{2} }[/tex]]
[tex]\frac{dV}{dP}[/tex] = - 0.00332 [[tex]\frac{m^{3} }{kPa}[/tex]]
Therefore,
(i) the value of [tex]\frac{dV}{dP}[/tex] when P = 50kPa is - 0.00332 [tex]\frac{m^{3} }{kPa}[/tex]
(ii) the meaning of the derivative [tex]\frac{dV}{dP}[/tex] is the rate of change of volume with pressure.
(iii) and the units are [tex]\frac{m^{3} }{kPa}[/tex]
