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Question

One kmol of hydrogen (molecular weight = 2 kg/kmol, specific heat ratio γ=1.4) at 1 bar, 300 K mixes with one kmol of nitrogen (molecular weight = 28 kg/kmol, specific heat ratio γ=1.4 ) at 1 bar, 300 K in an adiabatic vessel. The final mixture is also at 1 bar, 300 K. The entropy change for the process is

A
-5.76 kJ/K
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B
0 kJ/K
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C
5.76 kJ/K
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D
11.53 kJ/K
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Solution

The correct option is D 11.53 kJ/KEntropy change can be calculated as, △s=n¯RΣXiln(Xi) where n = Number of moles ¯R = universal gas constant X1 = Mole friction of 1st gas =−2×8.314×[12ln(12)+12ln(12)] =−2×8.314×ln(12)=11.53kJ/K

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