Question

One kmol of hydrogen (molecular weight = 2 kg/kmol, specific heat ratio $\gamma =1.4$) at 1 bar, 300 K mixes with one kmol of nitrogen (molecular weight = 28 kg/kmol, specific heat ratio $\gamma =1.4$ ) 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
• B. 0 kJ/K
• C. 5.76 kJ/K
• D. 11.53 kJ/K

Answer 1

The correct option is D 11.53 kJ/K

Entropy change can be calculated as,

$△s=n\overline{R}\Sigma X_{i}ln\left(X_i\right)$

where n = Number of moles

$\overline{R}$ = universal gas constant

$X_1$ = Mole friction of 1st gas

$=-2\times 8.314\times [\frac{1}{2}ln\left(\frac{1}{2}\right)+\frac{1}{2}ln\left(\frac{1}{2}\right)]$

$=-2\times 8.314\times ln\left(\frac{1}{2}\right)=11.53kJ/K$