The internal energy \( U \) of an ideal gas is given by:

\[
U = n \cdot C_V \cdot T
\]

For a diatomic gas like hydrogen (\( \text{H}_2 \)), \( C_V = \frac{5}{2} R \), and for a monoatomic gas like helium (\( \text{He} \)), \( C_V = \frac{3}{2} R \).

Given that the internal energy of \( n_1 \) moles of hydrogen at temperature \( T \) is equal to the internal energy of \( n_2 \) moles of helium at temperature \( 2T \), we have:

\[
n_1 \cdot \frac{5}{2} R \cdot T = n_2 \cdot \frac{3}{2} R \cdot (2T)
\]

Simplifying:

\[
\frac{5}{2} n_1 = 3 n_2
\]

Rearrange to find the ratio \( \frac{n_1}{n_2} \):

\[
\frac{n_1}{n_2} = \frac{3}{5} \cdot \frac{2}{1} = \frac{6}{5}
\]

Answer: The ratio \( \frac{n_1}{n_2} \) is \( \frac{6}{5} \).

The root mean square (r.m.s.) velocity \( v_{\text{rms}} \) of gas molecules is given by:

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

where \( m \) is the molar mass of the gas. Since all gases are at the same temperature, \( v_{\text{rms}} \) is inversely proportional to the square root of their molar mass:

\[
v_{\text{rms}} \propto \frac{1}{\sqrt{m}}
\]

Among air, oxygen (\( \text{O}_2 \)), carbon dioxide (\( \text{CO}_2 \)), and hydrogen (\( \text{H}_2 \)), hydrogen has the lowest molar mass. Therefore, it has the highest r.m.s. velocity.

Answer: Hydrogen has the maximum r.m.s. velocity.

The root mean square (r.m.s.) velocity \( v_{\text{rms}} \) of gas molecules is given by:

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

For nitrogen (\( \text{N}_2 \)) and oxygen (\( \text{O}_2 \)) gases to have the same \( v_{\text{rms}} \):

\[
\sqrt{\frac{3 k_B T_{\text{N}_2}}{M_{\text{N}_2}}} = \sqrt{\frac{3 k_B T_{\text{O}_2}}{M_{\text{O}_2}}}
\]

Squaring both sides and simplifying:

\[
\frac{T_{\text{N}_2}}{T_{\text{O}_2}} = \frac{M_{\text{N}_2}}{M_{\text{O}_2}}
\]

Given:
- \( T_{\text{O}_2} = 127^\circ \text{C} = 127 + 273 = 400 \, \text{K} \),
- Molar masses \( M_{\text{N}_2} = 28 \) and \( M_{\text{O}_2} = 32 \).

Substitute values:

\[
T_{\text{N}_2} = \frac{28}{32} \times 400 = 350 \, \text{K}
\]

Convert \( 350 \, \text{K} \) to Celsius:

\[
350 - 273 = 77^\circ \text{C}
\]

Answer: The temperature is \( 77^\circ \text{C} \).

At Normal Temperature and Pressure (NTP), 1 mole of any ideal gas occupies 22.4 liters and contains Avogadro's number (\(6.022 \times 10^{23}\)) of molecules.

Since each gas sample has the same volume (1 cm³) and is at NTP, they all contain the same number of molecules. This is because, at the same conditions of temperature and pressure, equal volumes of gases have equal numbers of molecules (Avogadro's Law).

Answer: All samples have the same number of molecules.

The relation between the root mean square (r.m.s.) velocity \( v_{\text{rms}} \) and the most probable velocity \( v_{\text{mp}} \) of a gas is derived from their respective formulas:

1. **Most probable velocity** \( v_{\text{mp}} \):
\[
v_{\text{mp}} = \sqrt{\frac{2 k_B T}{m}}
\]

2. **Root mean square velocity** \( v_{\text{rms}} \):
\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

Dividing \( v_{\text{rms}} \) by \( v_{\text{mp}} \):

\[
\frac{v_{\text{rms}}}{v_{\text{mp}}} = \sqrt{\frac{3}{2}}
\]

Thus:

\[
v_{\text{rms}} = \sqrt{\frac{3}{2}} \, v_{\text{mp}}
\]

The root mean square (r.m.s.) speed \( v_{\text{rms}} \) of gas molecules is given by:

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

Since \( v_{\text{rms}} \propto \sqrt{T} \), if the temperature changes, the r.m.s. speed changes proportionally to the square root of the temperature.

Let the initial r.m.s. speed at \( T = 120 \, \text{K} \) be \( v \). Then, at \( T = 480 \, \text{K} \), the r.m.s. speed \( v' \) is:

\[
v' = v \cdot \sqrt{\frac{480}{120}} = v \cdot \sqrt{4} = 2v
\]

So, the r.m.s. speed at \( 480 \, \text{K} \) will be \( 2v \).

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.

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 \( PV = \frac{15}{28} RT \).