A) 3/2
B) 23/15
C) 35/23
D) 4/3
Correct Answer: A
Solution :
Mayor's formula is \[{{C}_{p}}-{{C}_{v}}=R\] and \[\gamma =\frac{{{C}_{p}}}{{{C}_{v}}}\] Therefore, using above two relations, we find \[{{C}_{V}}=\frac{R}{\gamma -1}\] For monoatomic gas; \[\gamma =\frac{5}{3}\] \[\therefore \] \[{{C}_{V}}=\frac{R}{(5/3)-1}=\frac{3}{2}R\] For diatomic gas; \[\gamma =\frac{7}{5}\] \[\therefore \] \[{{C}_{V}}=\frac{R}{(7/5)-1}=\frac{5}{2}R\] When these two moles are mixed, then heat required to raise the temperature to\[1{}^\circ C\]is \[{{C}_{V}}=\frac{3}{2}R+\frac{5}{2}R=4R\] Hence, for one mole, heat required \[=\frac{4R}{2}=2R\] \[\therefore \] \[{{C}_{v}}=2R\] \[\Rightarrow \] \[\frac{R}{\gamma -1}=2R\] \[\Rightarrow \] \[\gamma =\frac{3}{2}\] Alternate Solution As, \[\frac{{{n}_{1}}+{{n}_{2}}}{\gamma -1}=\frac{{{n}_{1}}}{{{\gamma }_{1}}-1}+\frac{{{n}_{2}}}{{{\gamma }_{2}}-1}\] Here, \[{{n}_{1}}=1,{{n}_{2}}=1,{{\gamma }_{1}}=\frac{5}{3},{{\gamma }_{2}}=\frac{7}{5}\] \[\therefore \] \[\frac{1+1}{\gamma -1}=\frac{1}{\left( \frac{5}{3} \right)-1}+\frac{1}{\left( \frac{7}{5} \right)-1}\] \[\Rightarrow \] \[\frac{2}{\gamma -1}=\frac{3}{2}+\frac{5}{2}\] \[\Rightarrow \] \[\frac{2}{\gamma -1}=\frac{8}{2}\] \[\Rightarrow \] \[\frac{2}{\gamma -1}=4\] \[\Rightarrow \] \[\gamma =\frac{2}{4}+1\] Hence, \[\gamma =\frac{3}{2}\]
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