2018 A/L Physics Past Paper · Question 06 (English Medium)

Answer: (4) a constant volume process.

What the question asks

We know: the gas is an ideal gas.

The increase of its internal energy equals the heat supplied: $\Delta U = \Delta Q$.

We want: the process where this happens.

Relevant theory

In a system, if the net exchange of heat is ∆Q, the change of internal energy is ∆U and the amount of work done is ∆W then, ∆ Q = ∆U + ∆W

But, when a gas in heated at constant volume all the heat supplied to it, is used to increase the temperature.

Required equations

So the gas must do no work.

Its volume must not change.

$\Delta Q = \Delta U + \Delta W$

$\Delta W = 0$

$\Delta W = p\,\Delta V = 0 \Rightarrow \Delta V = 0$

Reading the options

Look at each process in turn.

Cyclic: $\Delta U = 0$.

Adiabatic: $\Delta Q = 0$.

Constant pressure: the gas does work, $\Delta W = p\Delta V$.

Constant volume: $\Delta W = 0$.

Isothermal: $\Delta U = 0$.

This option says: a cyclic process.

In a cycle the gas comes back to where it started. So $\Delta U = 0$.

But here heat is supplied, and $\Delta U = \Delta Q$.

So it is wrong.

This option says: an adiabatic process.

In an adiabatic process no heat goes in or out. So $\Delta Q = 0$.

But here heat is supplied to the gas.

This option says: a constant pressure process.

At constant pressure the gas expands. It does work, $p\Delta V$.

So part of the heat becomes work. $\Delta U < \Delta Q$.

This option says: a constant volume process.

The volume does not change. So the gas does no work: $\Delta W = 0$.

Then all the heat raises the internal energy: $\Delta U = \Delta Q$.

So this is the answer.

This option says: an isothermal process.

The internal energy of an ideal gas depends only on its temperature.

Here the temperature does not change. So $\Delta U = 0$. All the heat becomes work.

Reasoning and calculation

Only one process has $\Delta W = 0$ while heat is supplied.

That is the constant volume process.

Checks

A quick check.

(1) and (5): $\Delta U = 0$.

(2): $\Delta Q = 0$.

(3): $\Delta W$ is not zero.

Only (4) is left.

Conclusion

$\Delta V = 0$

If $\Delta U = \Delta Q$, the gas does no work.

$$\Delta W = 0$$

So the volume does not change.

The answer is option (4).

Write the first law.

$$\Delta Q = \Delta U + \Delta W$$

Then ask which term is zero.

Exam tip

Explore

Heat is supplied to the gas in a cylinder.

If the piston can move, the gas pushes it up. Part of the heat becomes work.

If the piston cannot move, the gas does no work.

Then all the heat raises the internal energy. So $\Delta U = \Delta Q$.