Constant Pressure Heating

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 C_p = 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 = mC_pΔT = 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: