Question

If the degrees of freedom of a gas molecule is 6, then the total internal energy of the gas molecule at a temperature of \( 47 \ , ^\circ\text{C} \) (in eV) is (Boltzmann constant \( = 1.38 \times 10^{-23} \ , \text{J K}^{-1} \))

• A. \( 414 \times 10^{-4} \)
• B. \( 828 \times 10^{-4} \)
• C. \( 927 \times 10^{-4} \)
• D. \( 572 \times 10^{-4} \)

Hint:

- Equipartition of Energy: Average energy per degree of freedom per molecule = \( \frac{1}{2}kT \). - Total internal energy per molecule for \(f\) degrees of freedom = \( \frac{f}{2}kT \). - Convert temperature from Celsius to Kelvin: \( T(K) = T(^\circ C) + 273.15 \) (often 273 is used). - Boltzmann constant \(k \approx 1.38 \times 10^{-23} \, \text{J/K}\). - Electron charge \(e \approx 1.602 \times 10^{-19} \, \text{C}\). - \(1 \, \text{eV} = e \times (1 \, \text{Volt}) \approx 1.602 \times 10^{-19} \, \text{J}\).

Answer 1

The Correct Option is B

According to the equipartition of energy, the average internal energy associated with each degree of freedom of a molecule is \( \frac{1}{2}kT \), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature in Kelvin.
Given degrees of freedom \( f = 6 \).
The total internal energy per molecule is \( U_{molecule} = f \times \frac{1}{2}kT = \frac{f}{2}kT \).
Temperature \( T_C = 47 \, ^\circ\text{C} \).
Convert to Kelvin: \( T_K = T_C + 273.
15 \).
Using 273 for simplicity if significant figures allow.
\( T_K = 47 + 273 = 320 \) K.
Boltzmann constant \( k_B = 1.
38 \times 10^{-23} \, \text{J K}^{-1} \).
(The problem uses 'k' for Boltzmann const) Internal energy per molecule in Joules: \[ U_{molecule} = \frac{6}{2} \times (1.
38 \times 10^{-23} \, \text{J K}^{-1}) \times (320 \, \text{K}) \] \[ = 3 \times 1.
38 \times 10^{-23} \times 320 \, \text{J} \] \[ = 3 \times 1.
38 \times 3.
2 \times 10^{-21} \, \text{J} \] \[ = 4.
14 \times 3.
2 \times 10^{-21} \, \text{J} \] \( 4.
14 \times 3.
2 = 13.
248 \).
\[ U_{molecule} = 13.
248 \times 10^{-21} \, \text{J} \] We need to convert this energy to electronvolts (eV).
Charge of an electron \( e = 1.
602 \times 10^{-19} \, \text{C} \).
\( 1 \, \text{eV} = 1.
602 \times 10^{-19} \, \text{J} \).
So, \( U_{molecule} (\text{in eV}) = \frac{13.
248 \times 10^{-21} \, \text{J}}{1.
602 \times 10^{-19} \, \text{J/eV}} \) \[ = \frac{13.
248}{1.
602} \times 10^{-21 - (-19)} \, \text{eV} = \frac{13.
248}{1.
602} \times 10^{-2} \, \text{eV} \] Calculation of \( \frac{13.
248}{1.
602} \): \( \frac{13.
248}{1.
602} \approx \frac{13.
25}{1.
6} \approx 8.
28125 \).
Let's use \(1.
602\): \(13.
248 / 1.
602 \approx 8.
26966\).
So, \( U_{molecule} \approx 8.
26966 \times 10^{-2} \, \text{eV} \).
This is \( 0.
0826966 \, \text{eV} \).
To match the options like \( X \times 10^{-4} \text{ eV} \): \( 0.
0826966 \, \text{eV} = 826.
966 \times 10^{-4} \, \text{eV} \).
This is approximately \( 827 \times 10^{-4} \, \text{eV} \).
Option (2) is \( 828 \times 10^{-4} \).
This is the closest.
The difference might be due to using \( T_K = 47 + 273.
15 = 320.
15 \) K.
If \( T_K = 320.
15 \): \( U_{molecule} = 3 \times 1.
38 \times 10^{-23} \times 320.
15 = 13.
25451 \times 10^{-21} \, \text{J} \).
\( U_{molecule} (\text{in eV}) = \frac{13.
25451 \times 10^{-21}}{1.
602 \times 10^{-19}} = \frac{13.
25451}{1.
602} \times 10^{-2} \approx 8.
2737 \times 10^{-2} \, \text{eV} \).
\( = 827.
37 \times 10^{-4} \, \text{eV} \).
Still closer to 827 than 828.
The rounding in options might be approximate, or a slightly different value of k or e was used by question setter.
However, \(828 \times 10^{-4}\) is the closest option.