A) \[\frac{3}{2}\]
B) \[\frac{4}{3}\]
C) \[\frac{6}{5}\]
D) \[\frac{8}{7}\]
Correct Answer: C
Solution :
Heat absorbed at constant volume \[H=n{{C}_{V}}\Delta T\] where,\[n=\]number of moles\[=\frac{m}{M}\] Heat absorbed by helium \[H=\frac{1}{4}\times \frac{3}{2}R(131-11)\] \[=\frac{1}{4}\times \frac{3}{2}R\times 120\] \[=45R\] [For the (monoatomic gas),\[{{C}_{V}}=\frac{3}{2}R\]] Heat absorbed by nitrogen \[{{H}_{N}}=\frac{7}{28}\times \frac{5}{2}R(71-11)\] \[=\frac{1}{4}\times \frac{5}{2}R\times 60\] \[=\frac{75}{2}R\] [For\[{{H}_{2}}\](diatomic gas)\[{{C}_{V}}=\frac{5}{2}R\]] \[\frac{H}{{{H}_{N}}}=\frac{45R}{(75/2)R}=\frac{6}{5}\]You need to login to perform this action.
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