A) 1.24
B) 1.58
C) 1.46
D) 1.60
Correct Answer: C
Solution :
\[{{(C\upsilon )}_{mono}}=\frac{3}{2}R,{{({{C}_{\upsilon }})}_{di}}=\frac{5}{2}R\] \[\therefore \] \[1\times \frac{3}{2}R\times \Delta T+2\times \frac{5}{2}R\times \Delta T=3{{C}_{\upsilon }}\Delta T\] \[\frac{3}{2}R+5R=3{{C}_{\upsilon }}\] \[\therefore \] \[{{C}_{\upsilon }}=\frac{13R}{6}\] Again\[{{({{C}_{p}})}_{mono}}=\frac{5}{2}R,{{({{C}_{p}})}_{di}}=\frac{7}{2}R\] \[\therefore \]\[1\times \frac{5}{2}R\times \Delta T+2\times \frac{7}{2}R\times \Delta T=3{{C}_{p}}\Delta T\] \[\frac{5}{2}R=7R=3{{C}_{p}}\] \[\frac{19R}{2}=3{{C}_{p}}\] \[{{C}_{p}}=\frac{19R}{6}\] Therefore, ratio of specific heats of mixture\[=\frac{{{C}_{p}}}{{{C}_{\upsilon }}}\] \[=\frac{19R}{6}\times \frac{6}{13R}\] \[=1.46\]You need to login to perform this action.
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