A) 6R/19
B) 13R/6
C) 6R/13
D) 19R/6
Correct Answer: D
Solution :
[d] \[\frac{{{M}_{{{N}_{2}}}}}{{{M}_{{{H}_{2}}}}}=\frac{14}{1}{{M}_{{{N}_{2}}}}=14m\text{ and }{{\text{M}}_{{{H}_{e}}}}=m\] \[{{C}_{p}}=\frac{{{n}_{1}}{{C}_{{{p}_{1}}}}+{{n}_{2}}{{C}_{{{p}_{2}}}}}{{{n}_{1}}+{{n}_{2}}}=\frac{\frac{14m}{28}\times \frac{7}{2}R+\frac{m}{4}\times \frac{5}{2}R}{\frac{14m}{28}+\frac{m}{4}}\] \[=\frac{2\times \frac{7}{2}R+\frac{5}{2}R}{2+1}=\frac{7R+\frac{5}{2}R}{3}=\frac{19R}{6}\]
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