Question

The total energy of a gas mixture of one mole of oxygen and 3 moles of argon at a temperature \( T \) by neglecting vibrational modes is:

• A. \( 5 \, {RT} \)
• B. \( \frac{7}{2} \, {RT} \)
• C. \( \frac{5}{2} \, {RT} \)
• D. \( 9 \, {RT} \)
• E. \( 7 \, {RT} \)

Hint:

For ideal gases, the total energy is calculated by the formula \( E = \frac{f}{2} n R T \), where \( f \) depends on the type of gas (monoatomic or diatomic). The total energy for a mixture is simply the sum of the individual energies for each component.

Answer 1

Answer: (E)

The total energy for an ideal gas is given by: \[ E = \frac{f}{2} n R T \] where: - \( f \) is the number of degrees of freedom of the gas molecules,
- \( n \) is the number of moles of the gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature.
For a diatomic molecule like oxygen (O\(_2\)):
- The number of degrees of freedom \( f = 5 \) (3 translational and 2 rotational, neglecting vibrational modes).
For a monoatomic molecule like argon (Ar):
- The number of degrees of freedom \( f = 3 \) (3 translational, no rotation in ideal cases for argon).
Now, calculate the total energy for the mixture:
- The energy for 1 mole of oxygen: \[ E_{{O}_2} = \frac{5}{2} \cdot 1 \cdot R T = \frac{5}{2} R T \] - The energy for 3 moles of argon: \[ E_{{Ar}} = \frac{3}{2} \cdot 3 \cdot R T = \frac{9}{2} R T \] The total energy for the mixture is the sum of the energies of oxygen and argon: \[ E_{{total}} = E_{{O}_2} + E_{{Ar}} = \frac{5}{2} R T + \frac{9}{2} R T = \frac{14}{2} R T = 7 R T \] Thus, the total energy of the gas mixture is: \[ \boxed{7 \, {RT}} \]