Question

A 56.0%/44.0% by volume propane/n-butane mixture is cooled from 375°C to 105°C. FInd the change in ethalpy of this mixture by following the steps below. Assume ideal behavior. a) a) What is the average molar mass of the mixture?

b) What is the change in specific enthalpy of pure propane from 375°C to 105°C?

c) What is the change in specific enthalpy of pure n-butane from 375°C to 105°C?

d) What is the change in specific enthalpy of the mixture from 375°C to 105°C?

e) What is the change in enthalpy of the mixture from 375°C to 105°C in kJ/kg?

Answer 1

To answer this we will Require Cp data

Cp/R = A +BT +CT2 + DT-2    Where T is in K from 298 oK

Name	Chemical formula	Tmax	CpigP298/R	A	B*103	C*106	D*10+5
Propane	C3H8	1500	9.011	1.213	28.785	-8.824	0
n-Butane	C410	1500	11.928	1.935	36.915	-11.402	0

a)For Gases mol percent is same as volume percent

So 56 % / 44 % volume percent will be same as mole percent

molar mass of butane = 58 g/mol

molar mass of propane = 44 g/mol

Mavg = Xbutane*Mbutane + Xpropane*Mpropane = (0.56*44) + (0.44*58) = 50.16 g/mol

Hence average molar mass of mixture is 50.16 g/mol

b) For this part we have to to integrate pure propane so for this part we are asked to basically find the specific enthalpy

We will integrate this from T1 = 375 oC = 648.15 K to T2 = 105 oC =378.15 K

The Cp function is mentioned above hence we will simply put in above equation and integrate it

H = - 30544.4 J/mol

c)

For this part we have to to integrate pure butane so for this part we are asked to basically find the specific enthalpy

We will integrate this from T1 = 375 oC = 648.15 K to T2 = 105 oC =378.15 K

The Cp function is mentioned above hence we will simply put in above equation and integrate it

H = - 39971.1 J/mol

d) Specific enthalpy of mixture = Xpropane*Hpropane +Xbutane*Mpropane = (0.56*(-30544.4)) + (0.44*(-39971.1) = -34692.148 J/mol

e) In terms of Kg we will multiply it with average molar mass

H = (-34692.148)*(50.16) =- 1740158.144 J/g = - 1740158.144 KJ/Kg