A gas mixture contains n₁ moles of a monoatomic gas and n₂ moles of gas of rigid diatomic molecules. Each molecule in monoatomic and diatomic gas has 3 and 5 degrees of freedom respectively. If the adiabatic exponent ( C_p C_V ) for this gas mixture is 1.5 , then the ratio n₁ n₂ will be

Step-by-step solution

For a gas mixture, C_V= n₁ C_ V₁ +n₂ C_ V₂ n₁+n₂ and C_p= n₁ C_ p₁ +n₂ C_ p₂ n₁+n₂ For monoatomic gas, n=n₁, C_ V₁ = 3 2 R, C_ p₁ = 5 2 R For diatomic gas, n=n₂, C_ V₂ = 5 2 R and C_ p₂ = 7 2 R Now given for gas mixture, gathered C_p C_V =1.5 So, 1.5= n₁ n₂ ( 5 2 )+ 7 2 n₁ n₂ ( 3 2 )+ 5 2 9 2 ( n₁ n₂ )+ 15 2 = ( n₁ n₂ ) 5+7 n₁ n₂ =1 gathered