Q.The partial pressure of ethane over a solution containing $6.56 \times 10^{-3}$ g of ethane is 1 bar. If the solution contains $5.00 \times 10^{-2}$ g of ethane, then what shall be the partial pressure of the gas?

By Henry's law, at constant temperature the mass of a gas dissolved in a fixed amount of solvent is proportional to its partial pressure, so

With $m_1 = 6.56 \times 10^{-3}\ \text{g}$, $p_1 = 1\ \text{bar}$ and $m_2 = 5.00 \times 10^{-2}\ \text{g}$: …

Step 1 — Apply Henry's law.

Henry's law states that, at a fixed temperature, the amount (mass) of a gas dissolved in a given quantity of solvent is directly proportional to the partial pressure of that gas above the solution:

Step 2 — Set up the ratio for the two cases.

Since the temperature and solvent are unchanged, the ratio $m/p$ is the same before and after:

$\frac{m_1}{p_1} = \frac{m_2}{p_2}$ …

Method: Direct Proportionality via Henry’s Law (Mass-Based Form)

Concept: Henry’s Law states that at constant temperature, the mass of a gas dissolved in a given volume of solvent is directly proportional to the partial pressure of the gas above the solution.

or

Steps

1. Identify given data

   • First case: $m_1 = 6.56 \times 10^{-3} \, \text{g}$, $P_1 = 1 \, \text{bar}$
   • Second case: $m_2 = 5.00 \times 10^{-2} \, \text{g}$, $P_2 = ?$

2. Apply Henry’s Law proportionality

   Since temperature and solvent are the same:

3. Rearrange for $P_2$

4. Substitute values $P_2 = 1 \, \text{bar} \times \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}}$ …

🔍 The Core Concept First

Henry’s Law states:

At constant temperature, the mass of a gas dissolved in a given volume of a liquid is directly proportional to the partial pressure of the gas above the liquid.

Mathematically:

or

Here:

• $m$ = mass of gas dissolved
• $P$ = partial pressure of the gas

Why this works: The amount of gas that dissolves depends on how many gas molecules are hitting the liquid surface — more pressure means more collisions, so more gas dissolves.

✗ Common Mistake #1: Using the wrong proportionality

The error: Students often think $P \propto \frac{1}{m}$ or use $P \propto m^2$, etc.

How to avoid:

Always write the proportionality clearly:

Tip: Remember — more pressure pushes more gas into solution, so mass and pressure go together, not opposite.

✗ Common Mistake #2: Forgetting to convert units

The error: Using grams and bars inconsistently, or mixing units like atm and bar without conversion.

How to avoid:

Check that both masses are in the same unit (here both are in grams — good). Also ensure pressure units match (here both in bar — fine). If not, convert before plugging in.

Tip: Always write units next to numbers in your working.

✗ Common Mistake #3: Plugging numbers without checking the logic

The error: Students directly substitute into $P_2 = \frac{m_2}{m_1} \times P_1$ but forget that $m_2 > m_1$ means $P_2 > P_1$.

How to avoid:

Before calculating, ask:

• Is the new mass larger or smaller?
• So should the new pressure be larger or smaller?

Here, $5.00 \times 10^{-2} > 6.56 \times 10^{-3}$, so $P_2 > 1$ bar. If you get a smaller number, you know something is wrong. …