A mixture of 2g H₂ and 32g O₂ — find partial pressure of each at 1atm

Question

A gaseous mixture contains 2 g of $\text{H}_2$ and 32 g of $\text{O}_2$. If the total pressure of the mixture is 1 atm, find the partial pressure of each gas.

Solution — Step by Step

Why This Works

Dalton’s Law of Partial Pressures states that in a mixture of non-reacting ideal gases, each gas behaves as if it were alone in the container. The pressure each gas exerts depends only on how many moles of it are present relative to the total.

The mole fraction is the key bridge: it tells us what fraction of the total pressure each component “owns.” If a gas makes up 30% of the moles, it contributes 30% of the total pressure.

Here, despite $\text{H}_2$ being much lighter (2 g/mol vs 32 g/mol), we had equal masses that happened to give equal moles. So both gases contribute exactly half the pressure. This is a common “trap” in exams — mass is not what matters for partial pressure; moles are.

Alternative Method — Direct from Ideal Gas Law

Since both gases are in the same container (same $V$, same $T$):

Since $n_{\text{H}_2} = n_{\text{O}_2}$, we immediately know $P_{\text{H}_2} = P_{\text{O}_2}$.

And since $P_{\text{H}_2} + P_{\text{O}_2} = 1$ atm, each must be 0.5 atm.

This shortcut works here because the moles are equal. Always check for such symmetry before doing full calculations.

Common Mistake