A container holds a mixture of 2 moles of a diatomic gas and 4 moles of a monatomic gas at a temperature T . The ratio of specific heats ( = C_p C_v ) for this gas mixture is:

Step-by-step solution

The molar heat capacity at constant volume for the mixture is given by: C_ v,mix = n₁ C_ v1 + n₂ C_ v2 n₁ + n₂ For the diatomic gas ( n₁ = 2 moles), C_ v1 = 5R 2 . For the monatomic gas ( n₂ = 4 moles), C_ v2 = 3R 2 . Substituting the values: C_ v,mix = 2 ( 5R 2 ) + 4 ( 3R 2 ) 2 + 4 = 5R + 6R 6 = 11R 6 The molar heat capacity at constant pressure for the mixture is: C_ p,mix = C_ v,mix + R = 11R 6 + R = 17R 6 The ratio of specific heats for the mixture is: = C_ p,mix C_ v,mix = 17R 6 11R 6 = 17 11 Answer: 17 11