A 46.0%/54.0% by volume propane/n-butane mixture is cooled from 255°C to 193°C.

Question

A 46.0%/54.0% by volume propane/n-butane mixture is cooled from 255°C to 193°C. FInd the change in ethalpy of this mixture by following the steps below. Assume ideal behavior. a) What is the average molar mass of the mixture? b) What is the change in specific enthalpy of pure propane from 255°C to 193°C? c) What is the change in specific enthalpy of pure n-butane from 255°C to 193°C? d) What is the change in specific enthalpy of the mixture from 255°C to 193°C? e) What is the change in enthalpy of the mixture from 255°C to 193°C in kJ/kg?

Explanation / Answer

Dear Student,

1. Average Molar Mass of the mixture is calculated as follows :

Molar mass of propane is 44.01 g/moles

Molar mass of n-butane is 58.12 g/moles

Therefore average molar mass of the mixture is (m) = [46 % x 44.01] + [ 54% x 58.12]

(m) = [0.46 x 44.01] + [0.54 x 58.12]

(m) = [20.24] + [31.38]

(m) = 20.24 + 31.38

(m) = 51.62

Hence the average molar mass of the mixture is 51.62

2. Change in specific enthalpy of pure propane from 2550 C to 193o C is calculated as follows:

Specific Enthalpy = H/ m

Here H is enthalpy and m is the mass

First we have to calculate H which is given by the formula H = Cp x dT

Here Cp of propane is 0.39 J/kg K and dT = 255-193 = 620 C = 273 + 62 = 335 K

H = Cp x dT = 0.39 x 335 = 130.65 J/kg

Specific Enthalpy change of propane = H / m = 130.65 / 0.044 = 2968.64 Joules

Therefore the change in specific enthalpy of pure propane is 2968.64 Joules = 2.968 kJ

3. Change in specific enthalpy of n-butane from 2550 C to 193o C is calculated as follows:

Specific Enthalpy = H/ m

Here H is enthalpy and m is the mass

First we have to calculate H which is given by the formula H = Cp x dT

Here Cp of n-butane is 0.39 J/kg K and dT = 255-193 = 620 C = 273 + 62 = 335 K

H = Cp x dT = 0.39 x 335 = 130.65 J/kg

Specific Enthalpy change of n-butane = H / m = 130.65 / 0.05812  = 2247.93 Joules

Therefore the change in specific enthalpy of n-butane is 2247.93 Joules = 2.247 kJ

4.What is the change in specific enthalpy of the mixture from 2550 C to 1930 C ?

The change in specific enthalpy of the mixture is given as =

change in specific enthalpy of pure propane + change in specific enthalpy of n-butane

Specific Enthalpy Change of the mixture = 2.968 + 2.247 = 5.215 kJ

Therefore, the change in specific enthalpy of the mixture is 5.215 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.