A stopcock that connects 2 bulbs of volumes 5 litres and 10 litres each containing the same ideal gas at 9 atm and 6 atm respectively is opened. What is the final pressure in the two bulbs if the temperature is maintained at a constant value?
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Solution

Equations: PV = nRT for an ideal gas.
Given: Two bulbs 1 and 2, both maintained at the same temperature (T). $P_{1} = 6 a t m, P_{2} = 9 a t m,$
$V_{1} = 10 L, V_{2} = 5 L$

Key idea: The total number of moles of the gas in both the bulbs remains a constant, even after opening the stop cock.

Calculations:
Using the ideal gas equation,

$n_{1} + n_{2} = n_{f i n a l}$
$\frac{P_{1} V_{1}}{R T} + \frac{P_{2} V_{2}}{R T} = \frac{P_{f} V_{f}}{R T}$ (Using Ideal gas law and the fact that temperature is constant)
$P_{1} V_{1} + P_{2} V_{2} = P_{f} V_{f}$
Except $P_{f} a n d V_{f}$ all other quantities are known.
$V_{f} = V_{1} + V_{2} = 10 + 5 = 15 L$ (Total volume will be sum of volumes of individual bulbs)
Substituting all values, $(6) (10) + (5) (9) = P_{f} (15)$
Solving $P_{f} = 7 a t m$

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