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Three moles of an ideal gas having gamma...

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.

A

`1.36`

B

`1.48`

C

`1.63`

D

`1.42`

Text Solution

Verified by Experts

The correct Answer is:
B
`C_("v, mix") = (n_1 C_(v_1) n_C_(V_2))/(n_1 + n_2) = (2 xx (3R)/2 + 3 xx (5R)/2)/(2 + 3) = (21R)/(10)`
`C_("p, mix") = C_("v, mix") + R = (31R)/(10) :. gamma_("mix") = (C_("p,mix"))/(C_("v,mix")) = 31/21 ~~ 1.48`.
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