A) \[P=2{{P}_{0}}\]
B) \[P=\frac{3}{4}{{P}_{0}}\]
C) \[2{{P}_{0}}{{V}_{0}}=3nR{{T}_{0}}\]
D) \[3{{P}_{0}}{{V}_{0}}=2nR{{T}_{0}}\]
Correct Answer: C
Solution :
\[{{P}_{0}}{{V}_{0}}={{n}_{0}}R{{T}_{0}}\][Initially for both containers] For container having temperature \[2{{T}_{0}},P{{V}_{0}}=nR\times 2{{T}_{0}}\] For container having temperature \[{{T}_{0}},P{{V}_{0}}=(2{{n}_{0}}-n)R{{T}_{0}}\] As number of moles of gas is conserved, so\[2{{n}_{0}}=3n\]\[\Rightarrow \]\[2\frac{{{P}_{0}}{{V}_{0}}}{R{{T}_{0}}}=3\times \frac{P{{V}_{0}}}{2R{{T}_{0}}}=3n\]\[\Rightarrow \]\[P=\frac{4}{3}{{P}_{0}}\]
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