A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is $3:2$. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of $ ext{Ne} = 20.2\ ext{u}$, mole

Question

A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is $3:2$. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of $	ext{Ne} = 20.2\ 	ext{u}$, molecular mass of $	ext{O}_2 = 32.0\ 	ext{u}$.

Accepted Answer

> [!TLDR]
> The key idea is that at the same temperature and volume, partial pressure is directly proportional to the number of moles (or molecules). Using the given pressure ratio $3:2$, we find the number ratio is also $3:2$. For mass density, we multiply the number ratio by the molecular mass ratio, giving $\frac{3 	imes 20.2}{2 	imes 32.0} = \frac{60.6}{64.0} \approx 0.947$.

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### Why partial pressure tells us the number of molecules

When two non-reactive gases share a vessel, they occupy the same volume $V$ and are at the same temperature $T$. The ideal gas law applies to each gas independently:

$$P_{	ext{Ne}} V = n_{	ext{Ne}} RT \quad 	ext{and} \quad P_{	ext{O}_2} V = n_{	ext{O}_2} RT$$

Here $n$ is the number of moles. Since $V$, $R$, and $T$ are identical for both gases, the partial pressure is directly proportional to the number of moles:

$$\frac{P_{	ext{Ne}}}{P_{	ext{O}_2}} = \frac{n_{	ext{Ne}}}{n_{	ext{O}_2}}$$

And because the number of molecules $N$ is just $n 	imes N_A$ (Avogadro’s number), the ratio of molecules is the same as the ratio of moles.

> [!TIP]
> This is a clean shortcut: for any mixture of ideal gases at the same $T$ and $V$, the partial pressure ratio equals the mole ratio *and* the molecule number ratio. No need to compute moles separately.

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### Step-by-step solution

**1. Ratio of number of molecules (i)**

Given $P_{	ext{Ne}} : P_{	ext{O}_2} = 3 : 2$, we have:

$$\frac{N_{	ext{Ne}}}{N_{	ext{O}_2}} = \frac{P_{	ext{Ne}}}{P_{	ext{O}_2}} = \frac{3}{2}$$

So the number of neon molecules is 1.5 times that of oxygen molecules.

> [!WARNING]
> A common mistake is to think the pressure ratio equals the mass ratio. It does not — pressure depends on the *number* of particles, not their mass. Oxygen molecules are heavier, so the mass ratio will be different.

**2. Ratio of mass densities (ii)**

Mass density $ho$ is mass per unit volume. For each gas:

$$ho_{	ext{Ne}} = \frac{m_{	ext{Ne}}}{V} = \frac{N_{	ext{N

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Q.A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is $3:2$ . Estimate the ratio of

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