Question

Two moles( n= 2) of an Idea gas with temperature T = 300K , P = 2bar and molar heat capacity Cvm = 1.5R are subjected consecutively to the following steps:

1) Gas is compressed Isothermally and reversibly to a pressure of 5bar

2) Following this the gas is expanded into vacuum until it volume reach V = 20 L

3) Finally there is a Isobaric change in temp to T = 350K

Question: Calculate the total heat exchanged during the whole process.

PLease write this out for me thanx!

Answer 1

1) Ideal gas law,

PV = nRT

V₁ = nRT / P₁ = (2 mol x 8.314 x 10^-2 L.bar/mol.K x 300 K) / 2 bar = 24.9 L

V₂ = nRT / P₂ = (2 mol x 8.314 x 10^-2 L.bar/mol.K x 300 K) / 5 bar = 9.9 L

Heat exchanged, Q = nRT ln (V₂/V₁) = 2 mol x 8.314 J/mol.K x 300 K x ln (9.9 L/24.9 L) = - 4600.97 J

Heat exchanged, Q = - 4600.97 J

2) Heat exchanged, Q = PΔV = 5 bar x (20 L - 9.9 L) = 5 bar x (10.1 L) = 50.5 bar.L = 5050.00 J

Heat exchanged, Q = 5050.00 J

3) Heat exchanged, Q = n C_vm ΔT = 2 mol x 1.5R x (350 K - 300K) = 2mol x 1.5 x 8.314 J/mol.K x 50 K = 1247.10 J

Heat exchanged, Q = 1247.10 J

Total Heat exchanged = - 4600.97 J + 5050.00 J + 1247.10 = 1696.13 J

Total Heat exchanged = 1696.13 J