Question

If the translational partition function for \(\mathrm{H_2}\) confined in a 1 L vessel at 300 K is \(y \times 10^{27}\), then the value of \(y\) is (rounded off to one decimal place).
(Given: Atomic mass (in amu): H = 1.008; 1 amu = \(1.661 \times 10^{-27}\) kg; \(h = 6.626 \times 10^{-34}\) J s; \(k = 1.381 \times 10^{-23}\) J K\(^{-1}\))

Hint:

Use \(q_{trans}=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V\) with \(m\) as the mass of one \(\mathrm{H_2}\) molecule in kg and \(V\) converted to \(\mathrm{m^3}\).

Answer 1

Correct Answer: 2.8

A cleaner way to get the same number is through the thermal de Broglie wavelength $\Lambda$, since $q_{trans} = V/\Lambda^3$ is just a repackaged form of the same formula.

1. Molecule mass: $m(\mathrm{H_2}) = 2(1.008)\ \text{amu} = 2.016\ \text{amu} = 2.016\times1.661\times10^{-27}\ \text{kg} = 3.3486\times10^{-27}\ \text{kg}$.
2. Thermal wavelength formula: $\Lambda = \dfrac{h}{\sqrt{2\pi m k T}}$. First get $2\pi m k T$ as before: $kT = 4.143\times10^{-21}\ \text{J}$, so $2\pi m kT = 8.7168\times10^{-47}$ (SI units).
3. Square root: $\sqrt{8.7168\times10^{-47}} = 9.3364\times10^{-24}$.
4. Wavelength: $\Lambda = \dfrac{6.626\times10^{-34}}{9.3364\times10^{-24}} = 7.0966\times10^{-11}\ \text{m}$.
5. Cube it: $\Lambda^3 = (7.0966\times10^{-11})^3 = 3.574\times10^{-31}\ \text{m}^3$.
6. Divide the volume: $V = 1\ \text{L} = 1\times10^{-3}\ \text{m}^3$, so $q_{trans} = \dfrac{1\times10^{-3}}{3.574\times10^{-31}} = 2.798\times10^{27}$.

Let's summarize:

• $q_{trans} = V/\Lambda^3$ is the same physics as the $(2\pi mkT/h^2)^{3/2}V$ formula, just grouped through the de Broglie wavelength.
• Both routes give the same number because they use the identical mass, temperature, and volume inputs.

So $q_{trans} = 2.80\times10^{27}$, meaning $\boxed{y = 2.8}$.