If one mole of a monoatomic gas ( = 5 3 ) is mixed with one mole diatomic gas ( = 7 5 ) , the value of for the mixture is

Step-by-step solution

For monoatomic gas, ₁= C_ p₁ C_ V₁ = 5 3 and C_ p₁ -C_ V₁ =R this gives C_ ₁ = 5 2 R and C_ V₁ = 3 2 R For diatomic gas, ₂= C_ ₂ C_ V₂ =7 / 5 and C_ p₂ -C_ V₂ =R C_ P₂ = 7 2 R and C_ V₂ = 5 2 R Now, we can write, _ mix = n₁ C_ p₁ +n₂ C_ ₂ n₁ C_ V₁ +n₂ C_ V₂ = C_ ₁ +C_ ₂ C_ V₁ +C_ V₂ = 5 2 R+ 7 2 R 3 2 R+ 5 2 R = 12 8 = 3 2 =1.5