Question

physical chemistry

4. Consider He to be a van der Waals gas. One mole of He is expanded adiabatically in a piston and cylinder from a volume of 1.0 l to 2.0 l. During the process, the temperature of the gas drops by 0.138 K (i.e. from 300.000 Kto 299.862 K). For He a = 0.0341 atm-e-/mole- and b=0.0238 l/mole. Find Cy for He.

Answer 1

He gas (vanderwaal) is given having n=1 mol undergoes adiabatie expansion V=1L to V₂ =22 AT= 0.38k ; T = Book, T2 =2990862k I

For adiabatic process; q=o : DU= qtW DU=w =-PdV = n(dt + na dv = = nRT dv + na dv V-nb (: P = _ART - na for vanderwaal) Cv dT = - R dv Doing integration of the equation; Crdt = -R 7 av V-nb JV-nb Cv la (72) = -R ln ( 16 nb) Ivi-nb

Oy 1 Putting values of every variable; b=0.0238 L/mol cv ln(299-862K) = -8.314 , X 3ook Jkt mot -- 10.0238 - mol L mor! 1L-1.0.02B mol Linot! Cv (-4.60106x104) = 8.314 X 0.7053 Jkt mot TC = 12 743.936 JK motil Thus, Cy value for He is 12743.936 Jktmolt