A) \[\frac{{{T}_{1}}{{T}_{2}}\left( {{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}} \right)}{{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}}\]
B) \[\frac{{{P}_{1}}{{V}_{1}}{{T}_{1}}+{{P}_{2}}{{V}_{2}}{{T}_{2}}}{{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}}\]
C) \[\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}+{{P}_{2}}{{V}_{2}}{{T}_{1}}}{{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}}\]
D) \[\frac{{{T}_{1}}{{T}_{2}}\left( {{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}} \right)}{{{P}_{1}}{{V}_{1}}{{T}_{1}}+{{P}_{2}}{{V}_{2}}{{T}_{2}}}\]
Correct Answer: A
Solution :
[a] Here \[Q=0\]and \[W=0.\] Therefore from first law of thermodynamics \[\Delta U=Q+W=0\] \[\therefore \]Internal energy of the system with partition = Internal energy of the system without partition. \[{{n}_{1}}{{C}_{v}}{{T}_{1}}+{{n}_{2}}{{C}_{v}}{{T}_{2}}=\left( {{n}_{1}}+{{n}_{2}} \right){{C}_{v}}T\] \[\therefore T=\frac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] \[\text{But }{{n}_{1}}=\frac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}\text{ and }{{n}_{2}}=\frac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}}\] \[\therefore T=\frac{\frac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}\times {{T}_{1}}+\frac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}}\times {{T}_{2}}}{\frac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}+\frac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}}}=\frac{{{T}_{1}}{{T}_{2}}\left( {{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}} \right)}{{{P}_{1}}{{V}_{1}}{{T}_{2}}+{{P}_{2}}{{V}_{2}}{{T}_{1}}}\]You need to login to perform this action.
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