2018 A/L Physics Past Paper · Question 39 (English Medium)

Biscuits in a closed container — humidity 80% → 30%

Biscuits of a freshly opened packet of biscuits were placed inside a container, and it was tightly closed with a lid, so that air cannot enter or leave the container. The initial relative humidity inside the container was also found to be 80%. After a few days, it was found that the relative humidity inside the container, had decreased to 30% and the mass of the biscuits had increased by $m$ . If the temperature inside the container remained constant throughout, the mass of water vapour initially present inside the container was

(1) $\dfrac{5m}{8}$
(2) $\dfrac{11m}{8}$
(3) $\dfrac{8m}{5}$
(4) $\dfrac{5m}{3}$
(5) $\dfrac{8m}{3}$

Answer and explanation

Answer: (3) $\dfrac{8m}{5}$

Given information

We know: the relative humidity falls from 80% to 30%.

The biscuits gain mass $m$ .

The container is closed. The temperature stays the same.

What the question asks

We want: the mass $m_0$ of vapour at the start, in terms of $m$ .

Relevant theory

The relative humidity of a given volume of air at some temperature is defined as the ratio of the mass ( m) of water vapour present in that volume to the mass ( m0) of water vapour present in the same volume at the same temperature when it is saturated with water vapour.

The temperature and the volume do not change.

So the vapour needed to saturate the air, $m_s$ , stays the same.

So the relative humidity is directly proportional to the mass of vapour.

Required equations

$RH = \frac{m_{vapour}}{m_s}$

$m = m_0 - m_{final}$

Applying it to this question

The plan.

Write the vapour at the start and at the end in terms of $m_s$ .

Their difference is $m$ . Use it to find $m_s$ .

Then find $m_0$ .

$0.80\,m_s$

$m_0 = 0.80\,m_s$

$0.30\,m_s$

$m_{final} = 0.30\,m_s$

Reasoning and calculation

$m = 0.80\,m_s - 0.30\,m_s$

$m = 0.50\,m_s$

$2m$

$m_s = \frac{m}{0.50}$

$m_0 = 0.80 \times 2m$

$\tfrac{8m}{5}$

$m_0 = 1.6m$

We got $m_0 = \frac{8m}{5}$ .

This matches option (3).

Checks

Check the unit.

$m_s = 2m$ and $m_0 = 1.6m$ . Both are masses.

The relative humidity is a ratio. It has no unit.

$0.8 * m / (0.8 - 0.3)$

Does it make sense?

$m_0 = 1.6m$ is more than $m$ . There was more vapour at the start than the biscuits took.

$0.6m$ is left at the end.

$1.6m - 0.6m = m$

Conclusion

$m_0 = \frac{8m}{5}$

The biscuits took $m = 0.5\,m_s$ .

So $m_s = 2m$ .

$m_0 = 0.8 \times 2m = \frac{8m}{5}$

The answer is option (3).

Explore

The biscuits are closed in the container. The relative humidity is 80%.

Over a few days the biscuits take in water vapour. Their mass grows by $m$ .

The relative humidity falls to 30%. So $m$ is 50% of the vapour that would saturate the air.

At the start there was 80% of it. That is $\frac{8m}{5}$ .

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