2022 A/L Physics Past Paper · Question 47 (English Medium)

Answer: (1) $\dfrac{0.9c - a}{0.9c}$, $\dfrac{a \times 100}{b}\%$

Given information

What we know:

Air at $30\ ^\circ\text{C}$ has a relative humidity of $90\%$.

The plant cools it to $10\ ^\circ\text{C}$ and removes some water. Then the air is heated to $20\ ^\circ\text{C}$.

The saturated vapour pressures are $a$ at $10\ ^\circ\text{C}$, $b$ at $20\ ^\circ\text{C}$ and $c$ at $30\ ^\circ\text{C}$.

There are no numbers, so we keep the letters. Later we check with sample values.

What the question asks

We want: the share of the moisture removed, and the final RH at $20\ ^\circ\text{C}$.

The options show the moisture removed as a fraction of the moisture that was in the air at the start.

What each letter means:

$a$, $b$, $c$ ($\text{Pa}$): the saturated water vapour pressure at $10\ ^\circ\text{C}$, $20\ ^\circ\text{C}$ and $30\ ^\circ\text{C}$.

$p$ ($\text{Pa}$): the vapour pressure of the water vapour in the air.

$RH$ ($\%$): the relative humidity, how full the air is of water vapour.

Relevant theory

What is happening:

At the start the air is not full of vapour. At $90\%$ RH its vapour pressure is $0.9c$.

The plant cools the air to $10\ ^\circ\text{C}$. Cold air can hold less vapour. So the vapour that does not fit turns into water and is taken out.

The air leaves the plant just full of vapour at $10\ ^\circ\text{C}$. Its vapour pressure is $a$.

Then the air is heated to $20\ ^\circ\text{C}$. Nothing is added or removed, so the vapour in the air stays as it was. But warm air can hold more, so the air is no longer full.

The options use only $a$, $b$ and $c$. So we take the air to stay at one total pressure, and the amount of vapour to go with its vapour pressure.

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.

p = vapour pressure at the given temperature

The other way is by reducing its temperature to reach the saturation point.

The first quote says that relative humidity is the vapour in the air compared with the vapour the same air would hold when full, at the same temperature.

The second quote names $p$ as the vapour pressure at the given temperature. Pressure and amount of vapour go together, so we can use pressures in place of amounts.

The third quote says that cooling is one way to make air full of vapour.

For example, a cold glass of water on a warm day. The air touching the glass is cooled. It cannot hold all its vapour, so tiny drops of water form on the outside of the glass.

Required equations

$p_1 = 0.9\,c$

$p_2 = a$

$\dfrac{p_1 - p_2}{p_1} = \dfrac{0.9c - a}{0.9c}$

$RH = \dfrac{a}{b} \times 100\%$

Applying it to this question

Every pressure in the fractions is in $\text{Pa}$, so each fraction has no unit. The share removed is a number between $0$ and $1$, and $RH$ is a percentage.

The cooling makes the air full only if $0.9c$ is bigger than $a$. It is, because air at $30\ ^\circ\text{C}$ can hold much more vapour than air at $10\ ^\circ\text{C}$.

$0.9c$

$p_1$

$a$

$p_2$

$\dfrac{p_1 - p_2}{p_1} = \dfrac{0.9c - a}{0.9c}$

$p_3$

$RH = \dfrac{a}{b} \times 100\%$

Reasoning and calculation

The share removed is $\dfrac{0.9c - a}{0.9c}$ and the final RH is $\dfrac{a \times 100}{b}\%$.

That is the pair in option (1).

Checks

$a * 100 / b$

Does it make sense?

With the sample values, about two thirds of the vapour is taken out and the air ends at about $52\%$ RH. Air that is $90\%$ full at $30\ ^\circ\text{C}$, cooled, dried and then warmed to $20\ ^\circ\text{C}$, feels drier.

The final RH is below $100\%$, as it must be for air that is not full.

The sample values are close to real ones but they are only for the check. The answer is in the letters.

Conclusion

$\dfrac{0.9c - a}{0.9c},\ \dfrac{a \times 100}{b}\%$

The share removed is $\dfrac{0.9c - a}{0.9c}$ and the final RH is $\dfrac{a \times 100}{b}\%$.

The answer is (1).

Write the vapour pressure at each stage. At the start it is $RH$ times the saturated pressure. If the cooling takes water out, the air is full, so it equals the saturated pressure at the cold temperature.

Heating without adding or removing vapour leaves the vapour pressure the same. Divide it by the saturated pressure at the new temperature for the new RH.

Exam tip