2022 A/L Physics Past Paper · Question 42 (English Medium)

Answer: (1) Graph (1)

What the question asks

What we know:

A fixed mass of an ideal gas goes from state $a$ to state $b$.

The figure shows the mean kinetic energy $E$ of the gas against its volume $V$. It is a straight line. Drawn backwards, it reaches the origin $0$. So $E$ is in proportion to $V$.

There are no numbers. We keep the letters, and later we check with sample values: $V = 2$ and $E = 6$ at $a$, $V = 5$ and $E = 15$ at $b$.

We want: the graph of pressure $P$ against volume $V$.

What each letter means:

$E$ ($\text{J}$): the mean kinetic energy of the gas.

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

$P$ ($\text{Pa}$): the pressure of the gas.

$T$ ($\text{K}$): the temperature of the gas on the kelvin scale.

$n$ ($\text{mol}$): the amount of gas, which is fixed.

$R$ ($\text{J mol}^{-1}\text{K}^{-1}$): the gas constant.

Relevant theory

What is happening:

The gas expands from $a$ to $b$. Its volume grows, and so does its mean kinetic energy.

The mean kinetic energy of a gas goes with its kelvin temperature. So $T$ grows in proportion to $E$.

The gas obeys $PV = nRT$. We use this to find $P$.

(as KE ∝ θ in kelvin)

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

The first quote is about a gas at constant temperature. It says $KE$ is in proportion to $\theta$ in kelvin, and that is the link we need: $E$ is in proportion to $T$.

For example, a party balloon left in the sun. The air inside gets warmer and the balloon swells. The soft rubber keeps the pressure inside close to the pressure of the air outside. So the volume grows in proportion to the temperature while the pressure stays about the same.

Required equations

$E \propto T$

$T \propto V$

$P = \dfrac{nRT}{V}$

$P = nR\dfrac{T}{V} = \text{a constant}$

Reasoning and calculation

$\text{mol} \times \text{J mol}^{-1}\text{K}^{-1} \times \text{K} \div \text{m}^3 = \text{J m}^{-3} = \text{Pa}$, so the result is a pressure.

$3$

$\dfrac{E}{V} = \dfrac{6}{2}$

$\dfrac{E}{V} = \dfrac{15}{5}$

$\dfrac{E}{V}$ is the same at $a$ and at $b$. Since $P$ goes with $\dfrac{T}{V}$, which goes with $\dfrac{E}{V}$, the pressure is the same at $a$ and at $b$.

A graph of $P$ against $V$ with the same pressure is a level line. That is graph (1).

Check the five graphs.

Graph (1) is level. The pressure is the same at $a$ and $b$.

Graphs (2) and (3) rise, and graphs (4) and (5) fall. In each of them the pressure at $b$ is different from the pressure at $a$.

Graph (1) matches.

Checks

$e / v$

Does it make sense?

The gas gets hotter as it expands. A gas that expands at a steady pressure does get hotter, like the balloon in the sun.

If the pressure fell as the gas expanded, $PV$ would grow more slowly than $V$, and $E$ would not stay in proportion to $V$. The $E$ graph here is a straight line through the origin, and that fits a level $P$ graph.

Conclusion

$\text{graph }(1)$

The pressure stays the same from $a$ to $b$. The graph is (1).

The answer is (1).

For an ideal gas, the mean kinetic energy goes with the kelvin temperature, and $PV = nRT$.

So $PV$ goes with $E$. If $E$ is in proportion to $V$, divide by $V$ and the pressure is a constant.

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