2019 A/L Physics Past Paper · Question 30 (English Medium)

Answer: (3) 720 J

Given information

We know:

$A$ has pressure $2 \times 10^4$ Pa and volume $3 \times 10^{-3}\ \text{m}^3$.

$B$ has pressure $6 \times 10^4$ Pa and volume $3 \times 10^{-3}\ \text{m}^3$.

$C$ has pressure $6 \times 10^4$ Pa and volume $6 \times 10^{-3}\ \text{m}^3$.

$D$ has pressure $2 \times 10^4$ Pa and volume $6 \times 10^{-3}\ \text{m}^3$.

The heat in $AB$ is $Q_{AB} = 200$ J. The heat in $BC$ is $Q_{BC} = 700$ J.

What the question asks

We want:

The change in internal energy $\Delta U_{ADC}$ along the path $ADC$.

Relevant theory

The first law:

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

The internal energy depends only on the start and the end. It does not depend on the path.

The work done by the gas is the area under the $P$–$V$ graph.

At constant pressure $W = P\Delta V$. At constant volume $W = 0$.

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

Required equations

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

$W = P\,\Delta V$

Applying it to this question

The plan.

$\Delta U$ does not depend on the path.

So find $\Delta U$ from $A$ to $C$ along $ABC$, where the heats are known.

That is also the answer for $ADC$.

Reasoning and calculation

$0$

$W_{AB} = 0$

$200$

$\Delta U_{AB} = 200 - 0$

$180$

$W_{BC} = 6 \times 10^4 \times (6 - 3) \times 10^{-3}$

$520$

$\Delta U_{BC} = 700 - 180$

$720$

$\Delta U_{AC} = 200 + 520$

$\Delta U_{ADC} = \Delta U_{ABC}$

Checks

Check the units.

$\text{Pa} \times \text{m}^3 = \text{N m}^{-2} \times \text{m}^3 = \text{N m} = \text{J}$.

Every work term is in joules.

Does it make sense?

On $ADC$ the work is only $W_{AD} = 2 \times 10^4 \times 3 \times 10^{-3} = 60$ J.

So the heat on this path is $720 + 60 = 780$ J.

That is less than the 900 J on $ABC$, because the gas does less work.

$\Delta U$ is 720 J on both paths.

$Qab + Qbc - P * dV$

Conclusion

$\Delta U = 720\ \text{J}$

$\Delta U_{AC} = 200 + (700 - 180) = 720$ J.

Internal energy does not depend on the path.

So it is also 720 J along $ADC$.

The answer is option (3).

Thinking that it cannot be found because the heat on the path $ADC$ is not given.

Internal energy depends only on the start and the end. The $\Delta U$ found on one path is true for every path.