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

Step 1: Write the translational partition function formula.
For a particle of mass \(m\) free to move in a volume \(V\), the translational partition function is
\[ q_{trans} = \left(\frac{2\pi m k T}{h^2}\right)^{3/2} V \] Here \(m\) is the mass of one molecule (not the molar mass), \(k\) is Boltzmann's constant, \(T\) is the temperature, and \(h\) is Planck's constant.

Step 2: Find the mass of one \(\mathrm{H_2}\) molecule.
Molar mass of \(\mathrm{H_2}\) in amu: \(2 \times 1.008 = 2.016\) amu. Convert to kg using \(1\ \mathrm{amu} = 1.661\times10^{-27}\) kg:
\[ m = 2.016 \times 1.661\times10^{-27}\ \mathrm{kg} = 3.3486\times10^{-27}\ \mathrm{kg} \]
Step 3: Compute \(2\pi m k T\).
\[ kT = (1.381\times10^{-23})(300) = 4.143\times10^{-21}\ \mathrm{J} \] \[ m\,kT = (3.3486\times10^{-27})(4.143\times10^{-21}) = 1.3873\times10^{-47} \] \[ 2\pi\,m\,kT = 6.2832 \times 1.3873\times10^{-47} = 8.7168\times10^{-47} \]
Step 4: Divide by \(h^2\).
\[ h^2 = (6.626\times10^{-34})^2 = 4.3904\times10^{-67} \] \[ \frac{2\pi m k T}{h^2} = \frac{8.7168\times10^{-47}}{4.3904\times10^{-67}} = 1.9854\times10^{20}\ \mathrm{m^{-2}} \]
Step 5: Raise to the power 3/2 and multiply by V.
\[ (1.9854\times10^{20})^{3/2} = (1.9854)^{3/2}\times10^{30} = 2.7976\times10^{30}\ \mathrm{m^{-3}} \] Convert \(V = 1\ \mathrm{L} = 1\times10^{-3}\ \mathrm{m^3}\):
\[ q_{trans} = 2.7976\times10^{30} \times 1\times10^{-3} = 2.7976\times10^{27} \]
Final Answer:
Comparing with \(q_{trans} = y \times 10^{27}\), and rounding to one decimal place, \[ \boxed{y \approx 2.8} \]