A) 1.50
B) 1.40
C) 1.53
D) 3.07
Correct Answer: A
Solution :
\[{{\gamma }_{1}}=5/3,\] gas is monoatomic \[{{C}_{v1}}\,=\frac{3R}{2}\] \[{{\gamma }_{2}}=7/5,\] gas is diatomic \[{{C}_{v2}}\,=\frac{5R}{2}\] \[{{C}_{v}}\,=\frac{{{n}_{1}}{{C}_{v1}}\,+{{n}_{2}}{{C}_{{{v}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}\] \[=\frac{(1)\left( \frac{3}{2}R \right)\,+(2)\,\left( \frac{5}{2}R \right)}{1+1}=2R\] \[{{C}_{P}}={{C}_{V}}+R=3R\] \[{{\gamma }_{mixture}}\,=\frac{{{C}_{p}}}{{{C}_{v}}}=1.5\]You need to login to perform this action.
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