2020 A/L Physics Past Paper · Question 13 (English Medium)

Answer: (2) $70\%$

Given information

System: two rooms, total volume $3V$, same temperature. Room A: $2V$, RH $60\%$; room B: $V$, RH $90\%$.

The door is left open for a long time — the vapour spreads evenly.

What the question asks

We must find: the final relative humidity common to both rooms after mixing.

Relevant theory

The relative humidity of a given volume of air at some temperature is defined as the ratio of the mass ( m) of water vapour present in that volume to the mass ( m0) of water vapour present in the same volume at the same temperature when it is saturated with water vapour.

Required equations

$m = RH \times m_0 = RH \times \rho_0 V$

$RH_A V_A + RH_B V_B = RH (V_A + V_B)$

Applying it to this question

What is conserved: the total mass of water vapour.

At one temperature the saturated vapour density $\rho_0$ is the same everywhere, so "$RH \times volume$" measures the vapour in each room.

Reasoning and calculation

Why it is conserved: the temperature does not change, and vapour neither condenses nor escapes — the door only lets it move between the two rooms.

$m_A = 0.60 \times \rho_0 (2V) = 1.2\,\rho_0 V$

$m_B = 0.90 \times \rho_0 V = 0.9\,\rho_0 V$

$m = RH \times \rho_0 (3V)$

$1.2 + 0.9 = RH \times 3$

$70$

$RH = \frac{2.1}{3}$

Checks

Plausibility: $70\%$ lies between $60\%$ and $90\%$, closer to the larger room A's value ✓.

Units: RH is a ratio — dimensionless ✓.

$(rhA * k + rhB) / (k + 1)$

Conclusion

$RH = \frac{0.6 \times 2V + 0.9V}{3V} = 70\%$

The total vapour does not change; it spreads through all $3V$: final RH $= 70\%$ — the answer is (2).

Taking the plain average of the two RH values ($75\%$).

Each RH must be weighted by its room's own volume — the larger room may hold more vapour.