Question

The specific heat at constant pressure of a gas is given by the following relation: Cp=0.85+0.00004T+5 x 10T2 where T is in Kelvin. Calculate the changes in enthalpy and internal energy of 10 kg of gas when its temperature is raised from 300 K to 2300 K. Take that the ratio of specific heats to be 1.5. A steel cylinder having a volume of 0.01653 m3 contains 5.6 kg of ethylene gas C2H4 molecular weight 28. Calculate the temperature to which the cylinder may be heated without the pressure exceeding 200 bar; given that compressibility factor Z = 0.605.

Answer 1

= (10/1.5)

= (10/1.5) × [0.85(2300 – 300) + (4 × 10-5/2)(23002 – 3002 ) + (5 × 10/3)(23003 – 3003)]

= 1.34 × 1012 KJ

Change in Internal Energy = 1.34 × 1012 KJ

Now; ν = 0.01653m3

Pν = ZRT

T = P.V/Z.R = [{200 × 105 × 0.01653}/{0.605 × (8.3143 × 103/28) }]

T = 1840.329K