A) \[1.40\]
B) \[1.50\]
C) \[1.53\]
D) \[3.07\]
Correct Answer: B
Solution :
Difference in gram-molecular heats \[({{C}_{P}}-{{C}_{V}})\] of an ideal gas is equal to gas constant\[R\]. \[\therefore \] \[{{C}_{P}}-{{C}_{V}}=R\] ... (i) (Mayors formula) Also the ratio of specific heat at constant pressure to specific heat at constant volume is \[\lambda =\frac{{{C}_{P}}}{{{C}_{V}}}\] ... (ii) From Eqs. (i) and (ii), we have \[{{C}_{V}}=\frac{R}{\gamma -1}\] For 1 mole of monoatomic gas \[{{C}_{V}}=\frac{R}{\frac{5}{3}-1}=\frac{3}{2}R\] For 1 mole of diatomic gas \[{{C}_{V}}=\frac{R}{\frac{7}{5}-1}=\frac{5}{2}R\] Thus, the heat required to raise the temperature of the mixture of \[1\] mole of monoatomic and \[1\] mole of diatomic gas by \[{{1}^{o}}C\] is \[\frac{3}{2}R+\frac{5}{2}R=4R\] Therefore, heat for \[1\] mole of the mixture to raise its temperature by \[{{1}^{o}}C\] is \[2R=\left( =\frac{4R}{2} \right)\], that is, for the mixture to-have \[i.e.,\] \[{{C}_{V}}=2R\] \[\frac{R}{\gamma -1}=2R\] or \[\gamma =1.50\] Alternative: \[\frac{{{n}_{1}}+{{n}_{2}}}{\gamma -1}=\frac{{{n}_{1}}}{{{\gamma }_{1}}-1}+\frac{{{n}_{2}}}{{{\gamma }_{2}}-1}\] For monoatomic gas,\[{{\gamma }_{2}}=\frac{7}{5},\,\,{{n}_{2}}=1\] \[\therefore \] \[\frac{1+1}{\gamma -1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}\]. \[\Rightarrow \] \[\frac{2}{\gamma -1}=\frac{3}{2}+\frac{5}{2}=4\] \[\Rightarrow \] \[\gamma -1=\frac{2}{4}=\frac{1}{2}\] \[\therefore \] \[\gamma =\frac{3}{2}=1.5\]
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