Q.34 A system consists of two reactors, connected by a valve. The first reactor (R1) contains
an ideal gas A of volume 5 L and the second reactor (R2) has an ideal gas B of volume 10 L.
Initially, the valve is closed and pressure P in R1 and R2 are 9 and 6 atm, respectively.
Later, when the valve is opened, the system reaches equilibrium. If the temperature T of both
the reactors is maintained constant, the final equilibrium pressure in atm of the system is _________.
Problem Breakdown
This GATE Biotechnology 2021 question tests ideal gas law application to find equilibrium pressure when gases mix at constant temperature. The correct final pressure is 7 atm, calculated using total moles conservation.
Two reactors connect via a valve:
- R1: 5 L at 9 atm containing gas A
- R2: 10 L at 6 atm containing gas B
- Temperature remains constant
After opening the valve, gases mix into total volume 15 L and reach equilibrium pressure.
Using Ideal Gas Law
For ideal gases at constant temperature:
PV = nRT ⇒ n ∝ PV
Initial moles:
R1: 5 × 9 = 45
R2: 10 × 6 = 60
Total moles:
45 + 60 = 105
Equilibrium Calculation
At equilibrium:
15P = 105
Therefore:
P = 105/15 = 7 atm
This aligns with Dalton’s law: total pressure equals the sum of partial pressures of mixed gases.
Alternate weighted formula:
P = (P1V1 + P2V2)/(V1 + V2) = (45 + 60)/15 = 7 atm
Common Mistakes Explained
- Taking simple average (9 + 6)/2 = 7.5 atm ignores volume effects
- Treating like Boyle’s law compression uses wrong constraints
- Ignoring constant T loses PV ∝ n logic
- Assuming reaction between gases—there is none
Why 7 atm is Correct
Check using partial pressures:
R1 contribution: 9 × 5 / 15 = 3 atm
R2 contribution: 6 × 10 / 15 = 4 atm
Total:
3 + 4 = 7 atm
