For a gas with \( \gamma = \frac{C_p}{C_v} = 1.4 \):

1. Atomicity: For diatomic gases, \( \gamma = 1.4 \) (common for diatomic molecules like \(\text{O}_2\), \(\text{N}_2\), etc.).

2. Heat capacities:
- \( C_v = \frac{R}{\gamma - 1} = \frac{R}{1.4 - 1} = \frac{5}{2} R \).
- \( C_p = \gamma C_v = 1.4 \times \frac{5}{2} R = \frac{7}{2} R \).

Thus, the atomicity is diatomic, and the values of \( C_p \) and \( C_v \) are \( \frac{7}{2} R \) and \( \frac{5}{2} R \), respectively.

To have the same rms speed for \(\text{H}_2\) and \(\text{N}_2\), we use the formula for rms speed:

\[
v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}
\]

Since the rms speeds are equal, we can set up the equation:

\[
\sqrt{\frac{3k_B T_{\text{H}_2}}{m_{\text{H}_2}}} = \sqrt{\frac{3k_B T_{\text{N}_2}}{m_{\text{N}_2}}}
\]

Square both sides and simplify:

\[
\frac{T_{\text{H}_2}}{m_{\text{H}_2}} = \frac{T_{\text{N}_2}}{m_{\text{N}_2}}
\]

Since \(\text{N}_2\) is 14 times heavier than \(\text{H}_2\), we have \( m_{\text{N}_2} = 14 \, m_{\text{H}_2} \) and \( T_{\text{N}_2} = 27^\circ \text{C} = 300 \, \text{K} \).

Now solve for \( T_{\text{H}_2} \):

\[
T_{\text{H}_2} = \frac{T_{\text{N}_2}}{14} = \frac{300}{14} \approx 21.4 \, \text{K}
\]

So, the temperature at which \(\text{H}_2\) has the same rms speed as \(\text{N}_2\) at 27°C is approximately \( 21.4 \, \text{K} \).

The root mean square (rms) speed \( v_{\text{rms}} \) of a gas is given by:

\[
v_{\text{rms}} = \sqrt{\frac{3RT}{M}}
\]

where \( M \) is the molar mass.

For helium (\( M = 4 \, \text{g/mol} \)) and argon (\( M = 40 \, \text{g/mol} \)) at the same temperature, the ratio of their rms speeds is:

\[
\frac{v_{\text{rms}(\text{He})}}{v_{\text{rms}(\text{Ar})}} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16
\]

Thus, the ratio \( \frac{v_{\text{rms}(\text{He})}}{v_{\text{rms}(\text{Ar})}} \) is approximately 3.16.

For a gas mixture of oxygen (\(\text{O}_2\)) and argon (\(\text{Ar}\)):

1. **Oxygen (\(\text{O}_2\))** is diatomic, so its internal energy per mole is:
\[
U_{\text{O}_2} = \frac{5}{2} RT
\]
For 2 moles, \( U_{\text{O}_2} = 2 \times \frac{5}{2} RT = 5 RT \).

2. **Argon (\(\text{Ar}\))** is monatomic, so its internal energy per mole is:
\[
U_{\text{Ar}} = \frac{3}{2} RT
\]
For 4 moles, \( U_{\text{Ar}} = 4 \times \frac{3}{2} RT = 6 RT \).

3. **Total internal energy**:
\[
U_{\text{total}} = U_{\text{O}_2} + U_{\text{Ar}} = 5 RT + 6 RT = 11 RT
\]

So, the total internal energy of the system is \( 11 RT \).

The mean kinetic energy of gas molecules depends only on temperature and is given by:

\[
\text{KE} = \frac{3}{2} k_B T
\]

where \( T \) is the temperature in Kelvin, and \( k_B \) is Boltzmann's constant.

Since we want the mean kinetic energy of \(\text{O}_2\) to equal that of \(\text{H}_2\) at \( -73^\circ \text{C} \):

1. Convert \(-73^\circ \text{C}\) to Kelvin:
\[
T_{\text{H}_2} = -73 + 273 = 200 \, \text{K}
\]

2. Since kinetic energy depends only on temperature, set the temperature \( T_{\text{O}_2} = T_{\text{H}_2} = 200 \, \text{K} \).

Therefore, the temperature at which \(\text{O}_2\) has the same mean kinetic energy as \(\text{H}_2\) at \(-73^\circ \text{C}\) is \(200 \, \text{K}\).

According to the kinetic theory of gases:

- (A) Collisions are always elastic: Gas molecule collisions do not lose kinetic energy, so they are elastic.
- (B) There is no force of attraction among the molecules: Assumption of ideal gases is no intermolecular forces.
- (C) Only a small number of molecules have very high velocity: Most molecules have moderate speeds; only a few have very high speeds.
- (D) Between collisions, the molecules move in straight lines with constant velocities: Molecules move with constant speed in straight lines until they collide.

All options are correct as per the assumptions of kinetic theory.

The root mean square (rms) velocity of a gas is given by:

\[
v_{\text{rms}} = \sqrt{\frac{3RT}{M}}
\]

where \( M \) is the molar mass. Since all gases are at the same temperature, the gas with the smallest molar mass will have the highest \( v_{\text{rms}} \).

Hydrogen has the smallest molar mass among the options, so it has the maximum rms velocity.

For 15 g of nitrogen gas (N₂), where the molar mass \( M = 28 \, \text{g/mol} \):

1. Calculate the number of moles, \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{15}{28} \).

2. Using the ideal gas law \( PV = nRT \):

\[
PV = \frac{15}{28} RT
\]

So, the equation of state is \( PV = \frac{15}{28} RT \).

For an ideal gas obeying \( PV^{\frac{1}{2}} = \text{constant} \):

1. Use the ideal gas law: \( PV = nRT \), so \( P = \frac{nRT}{V} \).
2. Substitute \( P = \frac{nRT}{V} \) in \( PV^{\frac{1}{2}} = \text{constant} \):

\[
\frac{nRT}{V} \cdot V^{\frac{1}{2}} = \text{constant} \Rightarrow nRT \cdot V^{-\frac{1}{2}} = \text{constant}
\]

3. At initial state, \( T = T \) and \( V = V \):

\[
T V^{-\frac{1}{2}} = \text{constant}
\]

4. When \( V \) changes to \( 4V \):

\[
T' (4V)^{-\frac{1}{2}} = T V^{-\frac{1}{2}}
\]

5. Simplify:

\[
T' \cdot \frac{1}{2} = T \Rightarrow T' = 2T
\]

So, the final temperature \( T' = 2T \).