(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$^{-1}$)
Calculating Translational Partition Function for H2
This solution details the calculation of the translational partition function ($q_{trans}$) for H2 gas to determine the value of y.
Step 1: Convert Given Values to SI Units
- Volume: $V = 1 \text{ L} = 1 \times 10^{-3} \text{ m}^3$
- Temperature: $T = 300 \text{ K}$
- Mass of H atom: $1.008 \text{ amu}$
- Mass of H2 molecule: $m = 2 \times 1.008 \text{ amu} = 2.016 \text{ amu}$
- Convert mass to kilograms: $m = 2.016 \text{ amu} \times 1.661 \times 10^{-27} \text{ kg/amu} = 3.349 \times 10^{-27} \text{ kg}$
- Planck's constant: $h = 6.626 \times 10^{-34} \text{ J s}$
- Boltzmann constant: $k = 1.381 \times 10^{-23} \text{ J K}^{-1}$
Step 2: Apply Translational Partition Function Formula
The formula for the translational partition function of a particle in a 3D box is:
$q_{trans} = \left(\frac{2 \pi m k T}{h^2}\right)^{3/2} V$Step 3: Calculate the Intermediate Term
Calculate the value inside the parenthesis:
$ \frac{2 \pi m k T}{h^2} = \frac{2 \pi \times (3.349 \times 10^{-27} \text{ kg}) \times (1.381 \times 10^{-23} \text{ J K}^{-1}) \times (300 \text{ K})}{(6.626 \times 10^{-34} \text{ J s})^2} $ $ \frac{2 \pi m k T}{h^2} = \frac{8.675 \times 10^{-47} \text{ kg J}}{4.390 \times 10^{-67} \text{ J}^2 \text{s}^2} \approx 1.976 \times 10^{20} \text{ m}^{-2} $Step 4: Compute the Translational Partition Function
Now, substitute this value back into the $q_{trans}$ formula:
$ q_{trans} = (1.976 \times 10^{20} \text{ m}^{-2})^{3/2} \times (1 \times 10^{-3} \text{ m}^3) $ $ q_{trans} = (1.976)^{1.5} \times (10^{20})^{1.5} \times 10^{-3} $ $ q_{trans} \approx 2.78 \times 10^{30} \times 10^{-3} $ $ q_{trans} \approx 2.78 \times 10^{27} $Step 5: Determine the Value of $y$
The question states that the translational partition function is $y \times 10^{27}$. By comparing this with the calculated value:
$y \times 10^{27} = 2.78 \times 10^{27}$ $y = 2.78$Rounding the value of y to one decimal place gives 2.8.
Important Questions from Partition Functions and Their Relation
Six distinguishable particles are distributed over 3 non‐degenerate levels, of energies 0, ε and 2ε. The most probable value for the total energy is
The partition function for a gas is given by
Q(N, V, T) = \(\frac{1}{N!}\left(\frac{2\pi m}{h^2\beta}\right)^{3N/2}\) (v - Nb)Ne \(\frac{\beta aN^2}{V}\)
The internal energy of the gas is
A three-state system with energies E = −ε0, 0, +ε0 is in a thermal equilibrium at a temperature T. If β ε0 = x, the probability of finding the system with energy E = 0 is [recall, cosh x = \(\frac{1}{2}\)(ex + e−x)]
The translational, vibrational, and rotational molecular partition functions for a system containing ideal diatomic gas molecules in the canonical ensemble (N, V, T) are written as, $q_{trans}$, $q_{vib}$, and $q_{rot}$, respectively. The option that correctly defines their thermodynamic variable(s) dependency is
If $q_t$ and $Q_{t,m}$ are the molecular and molar translational partition functions of $X_2$, respectively, then $ln(Q_{t,m})$ =
(N is the Avogadro number)
