A) 21.6 J/mol K
B) 19.7 J/mol K
C) 17.4 J/mol K
D) 15.7 J/mol K
Correct Answer: C
Solution :
\[{{f}_{mix}}=\frac{{{n}_{1}}{{f}_{1}}+{{n}_{2}}{{f}_{2}}}{{{n}_{1}}+{{n}_{2}}}=\frac{2\times 3+3\times 5}{5}=\frac{21}{5}\] \[{{C}_{\text{v}}}=\frac{fR}{2}=\frac{21}{5}\times \frac{R}{2}=17.4J/molK\]You need to login to perform this action.
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