A) Adiabatic constant of the gas mixture is 1.6
B) The average kinetic energy of the gas mixture after compression is in between \[18\text{R}{{\text{T}}_{0}}\] and \[19\text{R}{{\text{T}}_{0.}}\]
C) The final pressure of the gas mixture after compression is in between \[9{{\text{P}}_{0}}\] and \[10{{\text{P}}_{0}}.\]
D) The work \[\left| \,\text{W}\, \right|\]done during the process is \[13\text{R}{{\text{T}}_{0}}.\]
Correct Answer: A , C , D
Solution :
| monoatomic \[\left. \begin{align} & \text{monoatomic}\,\,\,\text{(5moles)} \\ & \,\,\,\,\text{diatomic}\,\,\,\,\,\,\,\,\,\,\,\text{(Imole)} \\ \end{align} \right\}{{\text{P}}_{0}},\,\,{{\text{V}}_{0,\,\,}}{{\text{T}}_{0}}\] |
| adiabtically compression to \[\frac{{{\text{V}}_{0}}}{4}\] |
| \[{{\gamma }_{mix}}=\frac{{{n}_{1}}{{C}_{p1}}+{{n}_{c}}{{C}_{{{p}_{z}}}}}{{{n}_{1}}{{C}_{{{Y}_{1}}}}+{{n}_{2}}{{C}_{{{V}_{2}}}}}=\frac{8}{5}\] |
| \[\text{W}=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{\gamma -1}\]. |
| and \[{{\text{p}}_{0}}{{\text{V}}_{0}}^{\text{8/5}}=\,{{\text{p}}_{2}}{{\left( \frac{{{\text{V}}_{0}}}{4} \right)}^{\text{8/5}}}\] |
| \[\therefore \] \[{{\text{P}}_{2}}=9.2{{\text{P}}_{0}}\] |
| \[\therefore \] \[\text{W}=\,\frac{{{\text{P}}_{0}}{{\text{V}}_{0}}-(9.2{{P}_{0}})({{\text{V}}_{0}}/4)}{3/5}\,=-13\text{R}{{\text{T}}_{0}}\] |
| \[\left| \text{W} \right|\,=\,13\text{R}{{\text{T}}_{0}}.\] |
Solution :
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