2020 A/L Physics Past Paper · Question 24 (English Medium)

Answer: (2) $5.4\ \text{kJ}$

Given information

From the graph: $a$: $V = 2 \times 10^{-3}\ \text{m}^3$, $p = 3 \times 10^5$ Pa; $b$: $V = 2 \times 10^{-3}$, $p = 8 \times 10^5$; $c$: $V = 5 \times 10^{-3}$, $p = 8 \times 10^5$.

Heat absorbed: $a \to b$ $6.0$ kJ, $b \to c$ $1.8$ kJ.

What the question asks

We must find: the change of internal energy $\Delta U_{ac}$ along the direct path $a \to c$.

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

∆ W = p.∆V (under constant pressure)

In an ideal gas internal energy appears only in the form of kinetic energy. (There are no intermolecular forces in a gas and hence there is no potential energy.)

∴ KE = 0 (as KE ∝ θ in kelvin)

Required equations

$\Delta U = \Delta Q - \Delta W$

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

$\Delta U_{ac} = \Delta U_{ab} + \Delta U_{bc}$

Applying it to this question

Plan: (1) $a \to b$ constant volume — $\Delta W = 0$, $\Delta U_{ab} = 6.0$ kJ.

(2) $b \to c$ constant pressure — $W = p\Delta V$, then $\Delta U_{bc}$.

(3) since an ideal gas's $U$ depends only on temperature (state), $\Delta U_{ac} = \Delta U_{ab} + \Delta U_{bc}$.

Reasoning and calculation

$6.0$

$\Delta U_{ab} = 6.0 - 0$

$2.4$

$W_{bc} = 8 \times 10^5 \times (5 - 2) \times 10^{-3}$

$-0.6$

$\Delta U_{bc} = 1.8 - 2.4$

$5.4$

$\Delta U_{ac} = 6.0 + (-0.6)$

Checks

Unit check: $\text{Pa} \times \text{m}^3 = \text{N m}^{-2} \times \text{m}^3 = \text{J}$; $10^5 \times 10^{-3} = 10^2$, so $8 \times 3 \times 10^2 = 2400$ J $= 2.4$ kJ ✓.

Physical plausibility: in $b \to c$, $p$ is constant while $V$ rises $2 \to 5$ — for an ideal gas, $T \propto pV$ ($16 \to 40$) rises.

So the internal energy must also rise ($\Delta U_{bc} > 0$): the heat absorbed should exceed the work done ($2.4$ kJ).

The given $1.8$ kJ contradicts this — the question's own data are impossible.

$Qab + Qbc - pb * (Vc - Va) * 0.1$

Conclusion

Official answer sheet: All — every option was accepted.

Reason: $b \to c$ is an expansion at constant pressure ($8 \times 10^5$ Pa). By the book's own sign convention, at constant pressure when heat is supplied $\Delta Q$, $\Delta U$ and $p\Delta V$ are all positive — the temperature rises. Here the work done by the gas is $p\Delta V = 2.4$ kJ, but the heat absorbed is only $1.8$ kJ — so $\Delta U_{bc} = -0.6$ kJ, i.e. the temperature should fall. But $T \propto pV$ rises. Both cannot be true together, so the question's data are faulty, and every candidate was given the mark.

What to take from this: the method is correct — apply $\Delta U = \Delta Q - \Delta W$ to each leg, and add the legs since $U$ depends only on state. Taking the data at face value, the answer is (2) $5.4$ kJ.

Teacher's note

$\Delta U_{ac} = 6.0 + 1.8 - 2.4 = 5.4\ \text{kJ}$

At face value, $\Delta U_{ac} = 6.0 - 0.6 = 5.4$ kJ — option (2). Since the data are contradictory, the official answer is: every option (All).

Adding the two absorbed heats ($7.8$ kJ) and taking that as the internal energy change.

The work $p\Delta V$ done by the gas as it expands must be subtracted: $\Delta U = \Delta Q - \Delta W$.