Question

A vessel contain 1 mole of o2an 2 mole of He. What is the value of 'Cp\Cv' of mixture

Answer 1

2 mol He+1 mol O2 We'll refer to He as gas 1 and O2 as gas 2 For a monatomic gas such as He: Cp1 = 5R/2 Cv1 = 3R/2 For a diatomic gas such as O2: Cp2 = 7R/2 Cv2 = 5R/2 Now for a mixture such as this the Cv value will depend on the mole fraction of each component such that: Cv = (n1Cv1 +n2Cv2)/(n1 + n2) Cv = (2Cv1+ Cv2)/(2 + 1) Cv = [(2 x 3R)/2 + (1 x 5R)2]/3 This simplifies down to Cv (mixture) = 33R/2 Now for the mixture : Cp (mixture) = Cv (mixture ) + R So Cp (mixture) = 33R/2 + R Cp (mixture) = 35R/2 Gamma = Cp(mixture)/Cv(mixture) = (35R/2)/(33R/2) = 35/33 = 1.06.