Question
2 MOLES OF MONOATOMIC GAS IS MIXED WITH 1 MOLE OF A DIATOMIC GAS. THEN γ FOR THE MIXTURE??
ANS 1.55
Joshi sir comment
total no of freedom for this mixture = (2*3+1*5)/(2+1) = 11/3
so γ = 1+[2/f] = 1 + 2/[11/3] = 1 + [6/11] = 17/11 = 1.55
2 MOLES OF MONOATOMIC GAS IS MIXED WITH 1 MOLE OF A DIATOMIC GAS. THEN γ FOR THE MIXTURE??
ANS 1.55
total no of freedom for this mixture = (2*3+1*5)/(2+1) = 11/3
so γ = 1+[2/f] = 1 + 2/[11/3] = 1 + [6/11] = 17/11 = 1.55