2019 A/L Physics Past Paper · Question 15 (English Medium)

Density of hydrogen from the root mean square speed

Hydrogen gas in a container is maintained at standard temperature (300 K) and pressure (1×105 N m−21 \times 10^5\ \text{N m}^{-2}). If the root mean square speed of hydrogen molecules is 2 km s−12\ \text{km s}^{-1}, what is the density of hydrogen in the container?

  1. (1) 0.038 kg m−30.038\ \text{kg m}^{-3}
  2. (2) 0.075 kg m−30.075\ \text{kg m}^{-3}
  3. (3) 0.150 kg m−30.150\ \text{kg m}^{-3}
  4. (4) 1.225 kg m−31.225\ \text{kg m}^{-3}
  5. (5) 2.450 kg m−32.450\ \text{kg m}^{-3}
Answer and explanation

Answer: (2) 0.075 kg m−30.075\ \text{kg m}^{-3}

Given information

We know:

The pressure p=1×105 N m−2p = 1 \times 10^5\ \text{N m}^{-2}.

The rms speed crms=2 km s−1=2000 m s−1c_{rms} = 2\ \text{km s}^{-1} = 2000\ \text{m s}^{-1}.

The temperature 300 K is only for a check.

What the question asks

We want:

The density of the hydrogen, ρ\rho, in kg m−3\text{kg m}^{-3}.

Relevant theory

The molecules hit the walls of the container.

Each hit changes the momentum of a molecule. This gives a force on the wall.

All the hits together give the pressure:

p=13ρc2‾p = \dfrac{1}{3}\rho\overline{c^2}

Here c2‾=crms2\overline{c^2} = c_{rms}^2.

In this theory, the following expression has been derived by considering the force due to the change of momentum resulted by the collisions of molecules with the walls of the container.

Required equations

p=13ρc2‾p = \frac{1}{3}\rho\overline{c^2}

ρ=3pcrms2\rho = \frac{3p}{c_{rms}^2}

Applying it to this question

Use SI units.

crms=2 km s−1=2×103 m s−1c_{rms} = 2\ \text{km s}^{-1} = 2 \times 10^3\ \text{m s}^{-1}.

So crms2=4×106 m2 s−2c_{rms}^2 = 4 \times 10^6\ \text{m}^2\text{ s}^{-2}.

Reasoning and calculation

ρ=3pcrms2\rho = \frac{3p}{c_{rms}^2}

ρ=3×1×105(2×103)2\rho = \frac{3 \times 1 \times 10^5}{(2 \times 10^3)^2}

ρ=3×1054×106\rho = \frac{3 \times 10^5}{4 \times 10^6}

0.0750.075

ρ=0.075\rho = 0.075

ρ=0.075 kg m−3\rho = 0.075\ \text{kg m}^{-3} matches option (2).

Checks

3∗p/(c∗c)3 * p / (c * c)

Check the units.

N m−2m2 s−2=kg m−1 s−2m2 s−2=kg m−3\dfrac{\text{N m}^{-2}}{\text{m}^2\text{ s}^{-2}} = \dfrac{\text{kg m}^{-1}\text{ s}^{-2}}{\text{m}^2\text{ s}^{-2}} = \text{kg m}^{-3}

That is the unit of density.

Does it make sense?

For hydrogen at 300 K:

crms=3RTM=3×8.31×3000.002≈1.93 km s−1c_{rms} = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3 \times 8.31 \times 300}{0.002}} \approx 1.93\ \text{km s}^{-1}

That agrees with the 2 km s−12\ \text{km s}^{-1} given.

Air at the same pressure has a density of about 1.2 kg m−31.2\ \text{kg m}^{-3}. Hydrogen is about 14 times lighter.

1.214≈0.08\dfrac{1.2}{14} \approx 0.08. That agrees too.

Conclusion

ρ=0.075 kg m−3\rho = 0.075\ \text{kg m}^{-3}

ρ=3pcrms2=3×1054×106=0.075 kg m−3\rho = \dfrac{3p}{c_{rms}^2} = \dfrac{3 \times 10^5}{4 \times 10^6} = 0.075\ \text{kg m}^{-3}

So the answer is option (2).

Change km s−1\text{km s}^{-1} into m s−1\text{m s}^{-1} before you square it.

(2 km s−1)2=4×106 m2 s−2(2\ \text{km s}^{-1})^2 = 4 \times 10^6\ \text{m}^2\text{ s}^{-2}.

Exam tip

தமிழ் · සිංහල