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Two moles of an ideal gas with Cₚ/C_v = 5/3 are mixed with 3 moles of another ideal gas with Cₚ/C_v = 4/3. The value of Cₚ/C_v for the mixture is:

1. 1.45
2. 1.50
3. 1.47
4. 1.42

Correct answer: 1.42

Solution

Cv1 = R/(5/3-1) = 1.5R, Cv2 = R/(4/3-1) = 3R. Cv_mix = (2*1.5R + 3*3R)/5 = 2.4R. Cp_mix = Cv_mix + R = 3.4R. So Cp/Cv = 3.4/2.4 = 1.42.

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