A) 3
B) 2
C) 4
D) 6
Correct Answer: B
Solution :
As for ideal gas \[{{C}_{P}}-{{C}_{V}}=R\] and \[\gamma =({{C}_{P}}/{{C}_{V}}),\] |
\[\gamma -1=\frac{R}{{{C}_{V}}}\] or \[{{C}_{V}}=\frac{R}{\left( \gamma -1 \right)}\]\[{{\left( {{C}_{V}} \right)}_{1}}=\frac{R}{\left( 5/3 \right)-1}=\frac{3}{2}R;\] |
\[{{\left( {{C}_{V}} \right)}_{2}}=\frac{R}{\left( 7/2 \right)-1}=\frac{5}{2}R\] |
and \[{{\left( {{C}_{V}} \right)}_{\operatorname{mix}}}=\frac{R}{\left( 19/13 \right)-1}=\frac{13}{6}R\] |
now from the conservation of energy, \[i.e.,\]\[\Delta U=\Delta {{U}_{1}}+\Delta {{U}_{2}},\]we get |
\[({{\mu }_{1}}+{{\mu }_{2}}){{({{C}_{V}})}_{\operatorname{mix}}}\Delta T=[{{\mu }_{1}}{{({{C}_{V}})}_{1}}+{{\mu }_{2}}{{({{C}_{V}})}_{2}}]\Delta T\] |
\[\Rightarrow \]\[{{({{C}_{V}})}_{\operatorname{mix}}}=\frac{{{\mu }_{1}}{{({{C}_{V}})}_{1}}+{{\mu }_{2}}{{({{C}_{V}})}_{2}}}{{{\mu }_{1}}+{{\mu }_{2}}}\] |
\[\frac{13}{6}R=\frac{1\times \frac{3}{2}R+n\times \frac{5}{2}R}{1+n}=\frac{(3+5n)}{2(1+n)}\] |
or \[13+13n=9+15n,\] i.e., number of gram mole, \[n=2\] |
You need to login to perform this action.
You will be redirected in
3 sec