To make the volume of an ideal gas four times its initial volume, we can use Boyle's Law, which states:

\[
P \propto \frac{1}{V} \quad \text{(at constant temperature)}
\]

If the initial pressure is \( P \) and the initial volume is \( V \), we want the final volume \( V_{\text{final}} = 4V \).

According to Boyle's Law:

\[
P_{\text{final}} \times V_{\text{final}} = P \times V
\]

Substituting \( V_{\text{final}} = 4V \):

\[
P_{\text{final}} \times 4V = P \times V
\]

\[
P_{\text{final}} = \frac{P}{4}
\]

So, to make the volume four times, we quarter the pressure at constant temperature.

To solve this, we can use the ideal gas law in terms of the number of moles \( n \):

\[
PV = nRT
\]

Let:
- Initial moles of hydrogen be \( n_1 \),
- Final moles of hydrogen be \( n_2 \).

Given data:
- Initial pressure \( P \),
- Final pressure \( P/2 \),
- Initial temperature \( T_1 = 500 \, \text{K} \),
- Final temperature \( T_2 = 300 \, \text{K} \),
- Mass of hydrogen initially = 6 g.

Since \( n = \frac{PV}{RT} \), we can write the initial and final moles as:

\[
n_1 = \frac{PV}{RT_1} \quad \text{and} \quad n_2 = \frac{(P/2)V}{R \cdot 300}
\]

Taking the ratio \( \frac{n_2}{n_1} \):

\[
\frac{n_2}{n_1} = \frac{(P/2) \cdot V / (R \cdot 300)}{P \cdot V / (R \cdot 500)} = \frac{1}{2} \times \frac{500}{300} = \frac{5}{12}
\]

Since initial moles \( n_1 = \frac{6}{2} = 3 \) moles (using molar mass of H₂ = 2 g/mol), then final moles \( n_2 = \frac{5}{12} \times 3 = 2.5 \) moles.

Thus, moles leaked out = \( 3 - 2.5 = 0.5 \) moles, corresponding to \( 0.5 \times 2 = 1 \) gram of hydrogen.

So, 1 g of hydrogen leaks out.

To find the number of molecules \( N \) of a gas, we can use the ideal gas law in terms of the number of molecules:

\[
PV = NkT
\]

where:
- \( P = 1.4 \times 10^7 \, \text{N/m}^2 \)
- \( V = 2 \times 10^{-3} \, \text{m}^3 \)
- \( T = 227^\circ \text{C} = 227 + 273 = 500 \, \text{K} \)
- \( k = 1.38 \times 10^{-23} \, \text{J/K} \) (Boltzmann constant)

Rearrange to solve for \( N \):

\[
N = \frac{PV}{kT}
\]

Substitute the values:

\[
N = \frac{(1.4 \times 10^7) \times (2 \times 10^{-3})}{(1.38 \times 10^{-23}) \times 500}
\]

Calculating this:

\[
N \approx 4.06 \times 10^{24}
\]

So, the number of molecules is \( 4.06 \times 10^{24} \).

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 \( v_{\text{rms}} \) for \( n \) molecules with speeds \( v_1, v_2, \ldots, v_n \) is calculated by:

\[
v_{\text{rms}} = \sqrt{\frac{v_1^2 + v_2^2 + v_3^2 + \cdots + v_n^2}{n}}
\]

Given speeds are: \( 2, 3, 4, 5, 6 \).

1. Calculate the squares:
\[
2^2 = 4, \quad 3^2 = 9, \quad 4^2 = 16, \quad 5^2 = 25, \quad 6^2 = 36
\]

2. Sum of the squares:
\[
4 + 9 + 16 + 25 + 36 = 90
\]

3. Divide by the number of molecules (5) and take the square root:
\[
v_{\text{rms}} = \sqrt{\frac{90}{5}} = \sqrt{18} \approx 4.24
\]

Thus, the r.m.s. speed is approximately \( 4.24 \).

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 the two rods to expand by the same length, their expansions \(\Delta l_1\) and \(\Delta l_2\) should be equal. The linear expansion for each rod can be written as:

\[
\Delta l_1 = l_1 \alpha_a t \quad \text{and} \quad \Delta l_2 = l_2 \alpha_s t
\]

Since \(\Delta l_1 = \Delta l_2\), we get:

\[
l_1 \alpha_a t = l_2 \alpha_s t
\]

Dividing both sides by \(t\) (assuming \(t \neq 0\)):

\[
l_1 \alpha_a = l_2 \alpha_s
\]

Now, to find the ratio \(\frac{l_1}{l_1 + l_2}\), divide both sides by \(\alpha_a + \alpha_s\):

\[
\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}
\]

So, the required ratio is:

\[
\frac{l_1}{l_1 + l_2} = \frac{\alpha_s}{\alpha_a + \alpha_s}
\]