Question

A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

• A. 29 RT
• B. 20 RT
• C. 27 RT
• D. 21 RT

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the total internal energy of a gas mixture consisting of argon and oxygen at temperature \(T\). The approach involves understanding the degrees of freedom for each type of molecule in the gas mixture.

Step 1: Identify the gases and their degrees of freedom

• Argon (Ar) is a monoatomic gas. For monoatomic gases, the degrees of freedom (\(f\)) are 3. The internal energy per mole for a monoatomic gas is given by:

\[\text{Internal energy per mole} = \frac{3}{2} RT\]

• Oxygen (O₂) is a diatomic gas. For diatomic gases, the degrees of freedom are 5 at ordinary temperatures, neglecting vibrational modes. The internal energy per mole for a diatomic gas is given by:

\[\text{Internal energy per mole} = \frac{5}{2} RT\]

Step 2: Calculate the total internal energy for each gas

• For argon, we have 8 moles: \(U_{\text{argon}} = 8 \times \frac{3}{2} RT = 12RT\)
• For oxygen, we have 6 moles: \(U_{\text{oxygen}} = 6 \times \frac{5}{2} RT = 15RT\)

Step 3: Calculate the total internal energy of the mixture

• Total internal energy: \(U_{\text{total}} = U_{\text{argon}} + U_{\text{oxygen}} = 12RT + 15RT = 27RT\)

The calculation matches with the given options, thus the total internal energy of the system is \(27 RT\). Therefore, the correct answer is 27 RT.

Answer 2

The total internal energy U of a gas mixture is given by:

\( U = nC_{V}T. \)

For argon (a monatomic gas), \( C_{V, Ar} = \frac{3R}{2} \). For oxygen (a diatomic gas), \( C_{V, O_2} = \frac{5R}{2} \).

Therefore, the internal energy of the mixture is:

\( U = n_1C_{V, Ar}T + n_2C_{V, O_2}T. \)

Substitute \( n_1 = 8 \), \( n_2 = 6 \):

\( U = 8 \times \frac{3R}{2} \times T + 6 \times \frac{5R}{2} \times T = 27RT. \)

Thus, the answer is:

\( 27RT. \)