A gaseous mixture consists of 4 ~g oxygen and 4 ~g of helium. The ratio C_p C_V of mixture is (C_p . and C_V are molar specific heats of the mixture at constant pressure and at constant volume respectively).

Step-by-step solution

For a gas mixture, r_ mixture = C_ p_ ( mix ) C_ V( mix ) = ( n₁ C_ p₁ +n₂ C_ p₂ n₁+n₂ ) ( n₁ C_ V₁ +n₂ C_ V₂ n₁+n₂ ) r_ mix = n₁ C_ p₁ +n₂ C_ p₂ n₁ C_ V₁ +n₂ C_ V₂ ...(i) Now, in given mixture we have 4 ~g of oxygen ( O ₂ ) and 4 ~g of helium (He). Now, number of moles of oxygen, n₁= m M = Sample mass Molar mass n₁= 4 32 = 1 8 Similarly number of moles of helium, n₂= 4 4 =1 Also oxygen is a diatomic gas so, C_ p₁ = 7 2 R and C_ V₁ = 5 2 R Helium is monoatomic, hence C_ p₂ = 5 2 R, C_ V₂ = 3 2 R Substituting these