Question

A gaseous mixture contains 2 moles of monatomic gas and 2 moles of diatomic gas at a temperature of 500 K. The total internal energy of the gaseous mixture is
(Atmospheric pressure = $10^5$ Pa and universal gas constant = $8.3 \, J mol^{-1} K^{-1}$)

• A. 28.6 kJ
• B. 24.8 kJ
• C. 33.2 kJ
• D. 27.2 kJ

Hint:

Remember degrees of freedom ($f$): Monatomic = 3 (translational only). Diatomic = 5 (3 translational + 2 rotational).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The internal energy ($U$) of an ideal gas depends on its temperature and degrees of freedom. For a mixture of non-reactive gases, the total internal energy is the sum of the internal energies of the individual gases. Formula: $U = \frac{f}{2} nRT$ Where $f$ is the degrees of freedom, $n$ is number of moles, $R$ is gas constant, $T$ is temperature.
Step 2: Key Values:

• Monatomic gas ($f_1 = 3$): $n_1 = 2$ moles.
• Diatomic gas ($f_2 = 5$ at moderate temperature): $n_2 = 2$ moles.
• $T = 500$ K.
• $R = 8.3 \, J mol^{-1} K^{-1}$.

Step 3: Calculation:
\[ U_{total} = U_{monatomic} + U_{diatomic} \] \[ U_{total} = \left( \frac{f_1}{2} n_1 RT \right) + \left( \frac{f_2}{2} n_2 RT \right) \] \[ U_{total} = \left( \frac{3}{2} \times 2 \times RT \right) + \left( \frac{5}{2} \times 2 \times RT \right) \] \[ U_{total} = 3RT + 5RT = 8RT \] Substitute values: \[ U_{total} = 8 \times 8.3 \times 500 \] \[ U_{total} = 4000 \times 8.3 \] \[ U_{total} = 33200 \, J = 33.2 \, kJ \]
Step 4: Final Answer:
The total internal energy is 33.2 kJ.