2023 A/L Physics Past Paper · Question 09 (English Medium)

Answer: (5) Graph (5)

What the question asks

We want: the graph of the density $\rho$ of the gas against its absolute temperature $T$, when the pressure stays the same.

What each letter means:

$\rho$ ($\text{kg m}^{-3}$): the density of the gas.

$T$ ($\text{K}$): the absolute temperature of the gas.

$m$ ($\text{kg}$): the mass of the gas. It is given and does not change.

$V$ ($\text{m}^3$): the volume of the gas.

$p$ ($\text{Pa}$): the pressure of the gas. It does not change.

There are no numbers. We keep the letters, and later we check with sample values: $300\ \text{K}$ at $a$ and $600\ \text{K}$ at $b$.

Relevant theory

What is happening:

The gas is warmed, and the pressure on it stays the same. So the gas spreads out and takes more room.

The mass of the gas stays the same. The same mass in a bigger volume is spread thinner, so its density is smaller.

We need to say how much smaller.

The volume of a fixed mass of a gas is directly proportional to its absolute temperature, provided the pressure remains constant.

The mass of a unit volume of a substance is termed its density.

The first quote is Charles' law. It says the volume is in proportion to the absolute temperature, when the pressure stays the same.

The second quote tells us what density is: mass in each unit of volume.

For example, think of a hot air balloon. The burner warms the air inside, and the air spreads out into a bigger space. The same air fills more room, so each cubic metre holds less of it. The balloon then floats up.

Required equations

$V = kT$

$\rho = \dfrac{m}{V}$

$\rho = \dfrac{m}{kT}$

$\rho \propto \dfrac{1}{T}$

Reasoning and calculation

$300$

$T_a$

$600$

$T_b = 2 \times 300$

$0.5$

$\dfrac{\rho_b}{\rho_a} = \dfrac{T_a}{T_b} = \dfrac{300}{600}$

When $T$ doubles, $\rho$ halves. It does not go to zero. It keeps getting smaller as $T$ grows.

Check the five graphs.

Graphs (1), (2) and (4) rise. A warmed gas gets thinner, so they cannot be right.

Graph (3) falls along a straight line. The density would fall by the same amount for each step of $T$, but it has to halve when $T$ doubles.

Graph (5) falls steeply at first and then slowly. It matches.

Checks

$t1 / t2$

Does it make sense?

A gas gets thinner when it is warmed at a steady pressure, like the air in the hot air balloon. So the density has to fall as $T$ rises.

The density has no way to become zero, because the gas is still there. So the graph gets closer and closer to the $T$ axis without touching it, and that is graph (5).

Conclusion

$\text{graph }(5)$

The density falls steeply at first and then slowly. The graph is (5).

The answer is (5).

For a gas at a steady pressure, the volume goes with $T$, and the density is mass over volume.

So the density goes with $\dfrac{1}{T}$. Look for the graph that falls steeply and then slowly.

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