A $22.4 L$ glass bulb has a total pressure of $760 t o r r$ at $0^{\circ} C$ and contains three different gases,
nitrogen, helium and argon. If the partial pressure of nitrogen is $250 t o r r$ and the partial pressure
of argon is $130 t o r r$ .
Given $R = 62.363 L . t o r r . m o l^{- 1} K^{- 1}$
Then the number of moles of helium in the bulb will be

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Solution

Given,
Total pressure = $760 t o r r$
Partial pressure of $N_{2}, p_{N_{2}} = 250 t o r r$
Partial pressure of $A r, p_{A r} = 130 t o r r$

Dalton’s law of partial pressures is a gas law which states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures exerted by each individual gas in the mixture.
Hence,
Total pressure = $p_{N_{2}} + p_{A r} + p_{H e_{2}}$
760 = 250 + 130 + $p_{H e_{2}}$
$p_{H e_{2}} = 760 - 380 = 380 t o r r$
The number of moles of He in the bulb is given by gas law,
PV = nRT
$R = 62.363 L . t o r r . m o l^{- 1} K^{- 1}$
$380 \times 22.4 = n_{H e} \times 62.363 \times 273$
$n_{H e} = 0.499 m o l \approx 0.5 m o l$

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