1.

Three moles of an ideal gas having `gamma = 1.67` are mixed with 2 moles of another ideal gas having `gamma = 1.4`. Find the equivalent value of `gamma` for the mixture.

Answer» Correct Answer - A::C
`(n_1 + n_2)/(gamma - 1)= (n_1)/(gamma_1 - 1)+(n_2)/(gamma_2 - 1)`
`:. (3 + 2)/(gamma -1) = 3/((1.67 - 1)) + 2/(1.4 - 1)`
`5/(gamma - 1) = 9.48`
Solving, we get `gamma = 1.53`.


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