Question

A vessel contains 3 moles of He, 1 mole of Ar, 5 moles of N$_2$ and 3 moles of H$_2$. If the vibrational modes are ignored, the total internal energy of the system of gases is

• A. 20 RT
• B. 26 RT
• C. 25 RT
• D. 30 RT

Hint:

Internal energy of $n$ moles of an ideal gas: $U = n \frac{f}{2} RT$.
Degrees of freedom ($f$):
Monatomic gas (He, Ar): $f=3$.
Diatomic gas (N$_2$, H$_2$) without vibrational modes: $f=5$. (3 translational, 2 rotational).
Total internal energy of a mixture is the sum of individual internal energies, assuming all gases are at the same temperature T.

Answer 1

The Correct Option is B

The internal energy ($U$) of $n$ moles of an ideal gas is given by $U = n C_V T$, where $C_V$ is the molar specific heat at constant volume. For an ideal gas, $C_V = \frac{f}{2}R$, where $f$ is the number of degrees of freedom and $R$ is the universal gas constant. So, $U = n \frac{f}{2} RT$. The total internal energy of a mixture of non-reacting gases is the sum of the internal energies of individual gases. All gases are assumed to be at the same temperature $T$. 1. Helium (He): $n_{He} = 3$ moles. He is a monatomic gas, so $f_{He} = 3$ (3 translational degrees of freedom). $U_{He} = n_{He} \frac{f_{He}}{2} RT = 3 \times \frac{3}{2} RT = \frac{9}{2} RT$. 2. Argon (Ar): $n_{Ar} = 1$ mole. Ar is a monatomic gas, so $f_{Ar} = 3$. $U_{Ar} = n_{Ar} \frac{f_{Ar}}{2} RT = 1 \times \frac{3}{2} RT = \frac{3}{2} RT$. 3. Nitrogen (N$_2$): $n_{N_2} = 5$ moles. N$_2$ is a diatomic gas. Ignoring vibrational modes, $f_{N_2} = 5$ (3 translational + 2 rotational). $U_{N_2} = n_{N_2} \frac{f_{N_2}}{2} RT = 5 \times \frac{5}{2} RT = \frac{25}{2} RT$. 4. Hydrogen (H$_2$): $n_{H_2} = 3$ moles. H$_2$ is a diatomic gas. Ignoring vibrational modes, $f_{H_2} = 5$. $U_{H_2} = n_{H_2} \frac{f_{H_2}}{2} RT = 3 \times \frac{5}{2} RT = \frac{15}{2} RT$. Total internal energy $U_{total} = U_{He} + U_{Ar} + U_{N_2} + U_{H_2}$: $U_{total} = \frac{9}{2} RT + \frac{3}{2} RT + \frac{25}{2} RT + \frac{15}{2} RT$. $U_{total} = \left(\frac{9+3+25+15}{2}\right) RT = \left(\frac{12+40}{2}\right) RT = \left(\frac{52}{2}\right) RT$. $U_{total} = 26 RT$. \[ \boxed{26 RT} \]