In given setup, container I has double the volume to that of container II. Container I & II are connected by a narrow tube with two knobs. 
Knob A : Closed – No gas is allowed to pass through
Open – All gases can pass
Knob B : Closed – A thin filter of Pd is introduced on complete cross section of tube
Open – All gases can pass.
Initially both knobs are closed. In container I, some amount of NH 3 gas is introduced which sets up equilibrium according to following reaction :
2NH 3 (g) 
N 2 (g) + H 2 (g)
Match the actions in column I to corresponding value in column II and select the correct answer using the code given below the column. Assume each action from initial stage.
Column I | Column II | ||
P. | A & B are closed, / = ? | 1. | 1/3 |
Q. | A is open & B is closed, / = ? | 2. | 3 |
R. | A is open & B is open, / = ? | 3. | 2 |
S. | A & B are left open for long time ; now B is closed & volume of container II is halved. (+)/(+) = ? | 4. | 2/3 |
Code
P Q R S
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(P) Initially, only NH 3 was present & according to reaction
2NH 3 (g) 
N 2 (g) + 3H 2 (g)
= 
Same volume for both gases ⇒ 
= 
⇒ 
= 3
(Q) When B is closed, only H 2 diffuses through filter until partial pressure of H 2 becomes equal.
= 3 ...
= 
& V I = 2V II ⇒ 
= 2 
+ 
= 
⇒ 
/ 
= 
...
from eq. &
= 
× 
= 
× 3 = 2
(R) Again same volume for N 2 & H 2
/ 
= 
...
= 
& V I = 2V II
⇒ 
= 2 
& 
= 2 
= 
+ 
= 3 
= 
= 
+ 
= 3 
from eq. 
/ 3 
= 
⇒ 
= 
(S) ( 
+ 
)/( 
+ 
) = 
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