Question
Two vessels containing gases $$A$$ and $$B$$ are interconected as shown in the figure. The stopper is opened, the gases are allowed to mix
homogeneously. The partial pressures of $$A$$ and $$B$$ in the mixture will be, respectively
Two vessels containing gases $$A$$ and $$B$$ are interconected as shown in the figure. The stopper is opened, the gases are allowed to mix
homogeneously. The partial pressures of $$A$$ and $$B$$ in the mixture will be, respectively
A.
8 and 5 atm
B.
9.6 and 4 atm
C.
4.8 and 2 atm
D.
6.4 and 4 atm
Answer :
4.8 and 2 atm
Solution :
$$\eqalign{ & {\text{Moles of}}\,A,\left( {{n_A}} \right) = \frac{{{P_A}{V_A}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 12}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{96}}{{RT}} \cr & {\text{Moles of}}\,B,\left( {{n_B}} \right) = \frac{{{P_B}{V_B}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 5}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40}}{{RT}} \cr} $$
$${\text{Total pressure}} \times {\text{total volume}}$$ $$ = \left( {{n_A} + {n_B}} \right) \times RT$$
$$\eqalign{ & p \times \left( {12 + 8} \right) = \frac{1}{{RT}}\left( {96 + 40} \right)RT \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = 6.8 \cr} $$
$${\text{Partial pressure of }}A$$ $$ = P \times {\text{mole fraction of }}A$$
$$\eqalign{ & = 6.8\left( {\frac{{\frac{{96}}{{RT}}}}{{\frac{{96 + 40}}{{RT}}}}} \right) \cr & = 4.8\,atm \cr} $$
$${\text{Partial pressure of}}\,B$$ $$ = 6.8 - 4.8 = 2\,atm.$$
$$\eqalign{ & {\text{Moles of}}\,A,\left( {{n_A}} \right) = \frac{{{P_A}{V_A}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 12}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{96}}{{RT}} \cr & {\text{Moles of}}\,B,\left( {{n_B}} \right) = \frac{{{P_B}{V_B}}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{8 \times 5}}{{RT}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40}}{{RT}} \cr} $$
$${\text{Total pressure}} \times {\text{total volume}}$$ $$ = \left( {{n_A} + {n_B}} \right) \times RT$$
$$\eqalign{ & p \times \left( {12 + 8} \right) = \frac{1}{{RT}}\left( {96 + 40} \right)RT \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P = 6.8 \cr} $$
$${\text{Partial pressure of }}A$$ $$ = P \times {\text{mole fraction of }}A$$
$$\eqalign{ & = 6.8\left( {\frac{{\frac{{96}}{{RT}}}}{{\frac{{96 + 40}}{{RT}}}}} \right) \cr & = 4.8\,atm \cr} $$
$${\text{Partial pressure of}}\,B$$ $$ = 6.8 - 4.8 = 2\,atm.$$