Constant Pressure Heating

Home -> ChE Learning Resources -> Solved Problems -> Thermodynamics->

One kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triples. Calculate W, Q, ΔU, and ΔH for the process. Assume air obeys the relation PV/T = 83.14 bar cm3 mol-1 K-1 and that Cp = 29 J mol-1 K-1.

Calculations:

From the relation, PV/T = 83.14 bar cm3 mol-1 K-1

Initial molal volume = 83.14 x 300 /1 = 24942 cm3/mol

= 24.942 m3/kmol

i.e. Initial volume of 1 kg of air (V1) = 24.942 m3 / 29 = 0.86 m3

Final volume (V2)= 3 x V1 = 3 x 0.86 = 2.58 m3 = 2.58 x 106 cm3.

Final specific volume = 24942 x 3 = 74826 cm3/mol

Final temperature:

P2 x 74826 = T2 x 83.14

1 x 74826 = T2 x 83.14

 T2 = 900 K

For the constant pressure heating process,

W = -P(V2 - V1) = -1 x (2.58 x 106 - 0.86 x 106) bar cm3 = -1 x 105 x (2.58 - 0.86) Nm = -172000 J = -172 kJ. (work is done by the system by expansion)

Moles of air in 1 kg mass = 1000/29 = 34.483 mol

Q = mCpΔT = 34.483 x 29 x (900 - 300) = 600000 J = 600 kJ.

ΔU = Q + W = 600 - 172 = 428 kJ (heat given to the system and work done on the system are positive quantities)

ΔH = Q = 600 kJ

Summary:

W

-172 kJ

Q

600 kJ

ΔU

428 kJ

ΔH

600 kJ
[Index]     [Learn More from Our Online Course...]

Last Modified on: 04-Feb-2022

Chemical Engineering Learning Resources - msubbu
e-mail: learn[AT]msubbu.academy
www.msubbu.in