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, \[{{\gamma }_{2}}=\frac{7}{2},{{n}_{2}}=1\] \[\therefore \] \[\frac{1+1}{\gamma -1}=\frac{1}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}\] \[\Rightarrow \] \[\frac{2}{\gamma -1}=\frac{3}{2}+\frac{5}{2}=4\] \[\Rightarrow \] \[\gamma -1=\frac{2}{4}=\frac{1}{2}\] \[\therefore \] \[\gamma =\frac{3}{2}=1.5\]
You need to login to perform this action.
You will be redirected in
3 sec
