Question

A gas in container A is in thermal equilibrium with another gas in container B both contains equal masses of the two gases in the respective containers. Which of the following can be true? (1) $ {{p}_{A}}{{V}_{A}}={{p}_{B}}{{V}_{B}} $ (2) $ {{p}_{A}}={{p}_{B}},{{V}_{A}}\ne {{V}_{B}} $ (3) $ \frac{{{p}_{A}}}{{{V}_{A}}}=\frac{{{p}_{B}}}{{{V}_{B}}} $ (4) $ {{p}_{A}}\ne {{p}_{B}},{{V}_{A}}={{V}_{B}} $

• A. 1, 2 and 3 are correct.
• B. 1 and 2 are correct.
• C. 2 and 4 are correct.
• D. 1 and 3 are correct.

Answer 1

Answer: (C)

According to problem mass of gases are equal so number of moles will not be equal ie, $ {{\mu }_{A}}\ne {{\mu }_{B}} $
From ideal gas equation $ pV=\mu RT $
$ \Rightarrow $ $ \frac{{{p}_{A}}{{V}_{A}}}{{{\mu }_{A}}}=\frac{{{p}_{B}}{{V}_{B}}}{{{\mu }_{B}}} $
[As temperature of the container are equal] From this relation it is clear that if
$ {{p}_{A}}={{p}_{B}}, $ then $ \frac{{{V}_{A}}}{{{V}_{B}}}=\frac{{{\mu }_{A}}}{{{\mu }_{B}}}\ne 1 $ ie, $ {{V}_{A}}\ne {{V}_{B}} $ Similarly, if $ {{V}_{A}}={{V}_{B}} $ then $ \frac{{{P}_{A}}}{{{P}_{B}}}=\frac{{{\mu }_{A}}}{{{\mu }_{B}}}\ne 1 $ ie, $ {{P}_{A}}\ne {{P}_{B}} $