2022 A/L Physics Past Paper · Question 37 (English Medium)

Answer: (4) $10\ \text{cm}$

Given information

What we know:

The sealed tube is $100\ \text{cm}$ long and lies flat. A mercury column $10\ \text{cm}$ long sits in the middle.

So the two air columns $P$ and $Q$ start with $\dfrac{100 - 10}{2} = 45\ \text{cm}$ each.

At the start both are at $27\ ^\circ\text{C}$ and at the same pressure.

Then $P$ is heated to $47\ ^\circ\text{C}$ and $Q$ to $127\ ^\circ\text{C}$.

What the question asks

We want: the difference between the lengths of the two air columns after heating.

What each letter means:

$p_0$ ($\text{Pa}$): the pressure of each air column at the start.

$p$ ($\text{Pa}$): the pressure of each air column at the end.

$x$ ($\text{cm}$): the final length of column $P$.

$y$ ($\text{cm}$): the final length of column $Q$.

$T_P$ ($\text{K}$): the final temperature of column $P$.

$T_Q$ ($\text{K}$): the final temperature of column $Q$.

Relevant theory

What is happening:

Heating makes the air in each column push harder, so both columns try to grow.

The mercury plug can slide along the tube. It stops moving when the air pushes on it equally from both sides.

So at the end the two columns have the same pressure $p$. The tube is flat, so the weight of the mercury does not push along the tube. We also neglect friction.

The air is trapped, so each column keeps its own amount of gas. The tube is the same width all along, so length stands for volume.

This relationship is known as the equation of state of a gas.

Using T = θ + 273.15

For one fixed amount of gas the equation of state says $\dfrac{pV}{T}$ stays the same. The temperature $T$ must be in kelvin. The book gives $T = \theta + 273.15$, and we round the $273.15$ to $273$.

For example, a closed bicycle tyre left in the hot sun gets harder. The air inside is warmer, so it pushes on the tyre with more force.

The two air columns are like two tyres of the same size. Each one gets warmer and pushes harder.

Required equations

$\dfrac{p\,x}{T_P} = \dfrac{p_0 \times 45}{300}$

$\dfrac{p\,y}{T_Q} = \dfrac{p_0 \times 45}{300}$

$\dfrac{x}{T_P} = \dfrac{y}{T_Q}$

$x + y = 90$

Applying it to this question

The pressures $p$ and $p_0$ cancel, so only lengths and temperatures remain. $\dfrac{\text{cm}}{\text{K}} = \dfrac{\text{cm}}{\text{K}}$ on both sides.

$320$

$T_P = 47 + 273$

$400$

$T_Q = 127 + 273$

$1.25$

$\dfrac{y}{x} = \dfrac{T_Q}{T_P} = \dfrac{400}{320}$

$40$

$x = 90 \times \dfrac{320}{320 + 400}$

$50$

$y = 90 \times \dfrac{400}{320 + 400}$

$10$

$y - x = 50 - 40$

Reasoning and calculation

The difference between the lengths is $10\ \text{cm}$.

That is option (4).

Checks

$a * (tq - tp) / (tq + tp)$

Does it make sense?

The hotter column $Q$ is longer, which is what we expect. The mercury has been pushed towards $P$, from $45\ \text{cm}$ to $40\ \text{cm}$ of air on that side.

Check the pressure: for column $P$ the equation of state gives $\dfrac{p \times 40}{320} = \dfrac{p_0 \times 45}{300}$, so $p = 1.2\,p_0$. For column $Q$ it gives $\dfrac{p \times 50}{400} = \dfrac{p_0 \times 45}{300}$, so $p = 1.2\,p_0$. Both columns have the same pressure, as the mercury needs.

Conclusion

$y - x = 10\ \text{cm}$

The lengths are $40\ \text{cm}$ and $50\ \text{cm}$. The difference is $10\ \text{cm}$.

The answer is (4).

When a free plug separates two gas columns, the pressure on its two sides is the same at the end. Use that with $\dfrac{pV}{T}$ for each column.

Before you put a temperature into a gas equation, change $^\circ\text{C}$ to kelvin.

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