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.) 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 \( 4.24 \).

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:

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

For a liquid in a U-tube with different temperatures in each arm, the expansion of the liquid in each arm is affected by the temperature difference.

Let:
- \( l_1 \) and \( l_2 \) be the heights of the liquid columns at temperatures \( t_1 \) and \( t_2 \), respectively.
- \( \beta \) be the coefficient of volume expansion of the liquid.

Since the pressure at the same horizontal level in both arms must be equal, we have:
\[
l_1 (1 + \beta t_1) = l_2 (1 + \beta t_2)
\]

Rearranging, we get:
\[
l_1 + l_1 \beta t_1 = l_2 + l_2 \beta t_2
\]

Solving for \( \beta \):
\[
\beta = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}
\]

Thus, the coefficient of volume expansion of the liquid is:
\[
\beta = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}
\]

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

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

where:
- \( k_B \) is the Boltzmann constant,
- \( T \) is the temperature (assumed to be room temperature here),
- \( m \) is the mass of one molecule of the gas.

Rearranging to find \( m \):

\[
m = \frac{3 k_B T}{v_{\text{rms}}^2}
\]

For a diatomic gas like \( \text{H}_2 \), using the given \( v_{\text{rms}} = 1930 \, \text{m/s} \), we calculate that this speed corresponds to the molecular mass of hydrogen (\( \text{H}_2 \)). Therefore, the gas is identified as \( \text{H}_2 \).

To find the ratio of the slopes of curves \( B \) to \( A \) for gases with masses \( m \) and \( 3m \) at constant volume:

1. For an ideal gas at constant volume, \( P = \frac{nRT}{V} \).
2. Since \( n = \frac{\text{mass}}{\text{molar mass}} \), the pressure \( P \propto \frac{\text{mass} \cdot T}{M} \).
3. So, slope \( \propto \frac{\text{mass}}{M} \) for each gas.

For gas \( A \) with mass \( m \), let the slope be \( S \propto \frac{m}{M} \).
For gas \( B \) with mass \( 3m \), slope \( S_B \propto \frac{3m}{M} \).

Thus, the ratio of slopes \( \frac{S_B}{S_A} = \frac{3m/M}{m/M} = 3:1 \).

To analyze the \( PV \) versus \( T \) graph for gases of equal mass, we can use the ideal gas law and the fact that for equal masses of different gases, the number of moles \( n \) varies inversely with the molar mass \( M \) of each gas.

1. **Ideal Gas Law**:
\[
PV = nRT
\]
For a given temperature \( T \), \( PV \propto n \). Thus, \( PV \) will be larger for gases with more moles (i.e., smaller molar mass).

2. **Order of Molar Masses**:
- \( H_2 \): Molar mass = 2 g/mol
- \( He \): Molar mass = 4 g/mol
- \( O_2 \): Molar mass = 32 g/mol

Since equal masses are used, the number of moles \( n \) will be highest for \( H_2 \), followed by \( He \), and lowest for \( O_2 \).

3. **Conclusion**:
- Line with the highest \( PV \) value corresponds to the gas with the highest number of moles, i.e., \( H_2 \) (line **C**).
- The middle line corresponds to \( He \) (line **B**).
- The line with the lowest \( PV \) value corresponds to \( O_2 \) (line **A**).

Thus, **C corresponds to \( H_2 \)**, **B to \( He \)**, and **A to \( O_2 \)**.

The total average kinetic energy \( E \) of translatory motion for an ideal gas is given by:

\[
E = \frac{3}{2} nRT
\]

where:
- \( n \) is the number of moles,
- \( R \) is the gas constant (\( 8.314 \, \text{J/mol·K} \)),
- \( T \) is the temperature in Kelvin.

We can rearrange to find \( nRT \):

\[
nRT = \frac{2}{3} E
\]

The ideal gas law also gives us:

\[
PV = nRT
\]

Thus,

\[
P = \frac{nRT}{V} = \frac{\frac{2}{3} E}{V}
\]

Substitute \( E = 1.5 \times 10^5 \, \text{J} \) and \( V = 20 \, \text{litres} = 20 \times 10^{-3} \, \text{m}^3 \):

\[
P = \frac{\frac{2}{3} \times 1.5 \times 10^5}{20 \times 10^{-3}}
\]

\[
P = \frac{10^5}{20 \times 10^{-3}} = 5 \times 10^6 \, \text{N/m}^2
\]

So, the pressure is \( 5 \times 10^6 \, \text{N/m}^2 \).

For a monoatomic gas under isobaric (constant pressure) conditions, the heat required \( Q \) is given by:

\[
Q = n C_p \Delta T
\]

where:
- \( n = 2 \) moles,
- \( C_p = \frac{5}{2} R \) (specific heat at constant pressure for monoatomic gas),
- \( \Delta T = 50^\circ \text{C} - 0^\circ \text{C} = 50 \) K.

Substitute values:

\[
Q = 2 \times \frac{5}{2} R \times 50 = 250R
\]

So, the answer is \( 250R \).

For a mixture of \( n_1 \) moles of a monatomic gas and \( n_2 \) moles of a diatomic gas, the ratio \( \gamma = \frac{C_p}{C_v} \) of the mixture is given by:

\[
\gamma = \frac{\text{Total } C_p}{\text{Total } C_v}
\]

1. Molar heat capacities:
- For a monatomic gas: \( C_{v, \text{mono}} = \frac{3}{2} R \) and \( C_{p, \text{mono}} = \frac{5}{2} R \).
- For a diatomic gas: \( C_{v, \text{di}} = \frac{5}{2} R \) and \( C_{p, \text{di}} = \frac{7}{2} R \).

2. Total heat capacities:
- Total \( C_v = n_1 \cdot \frac{3}{2} R + n_2 \cdot \frac{5}{2} R \).
- Total \( C_p = n_1 \cdot \frac{5}{2} R + n_2 \cdot \frac{7}{2} R \).

3. Given \( \gamma = 1.5 \):

\[
\frac{C_p}{C_v} = \frac{n_1 \cdot \frac{5}{2} R + n_2 \cdot \frac{7}{2} R}{n_1 \cdot \frac{3}{2} R + n_2 \cdot \frac{5}{2} R} = 1.5
\]

4. Simplify by canceling \( R \) and multiplying through by 2:

\[
\frac{5n_1 + 7n_2}{3n_1 + 5n_2} = 1.5
\]

5. Cross-multiply to solve for \( n_1 \) in terms of \( n_2 \):

\[
5n_1 + 7n_2 = 1.5 (3n_1 + 5n_2)
\]

\[
5n_1 + 7n_2 = 4.5n_1 + 7.5n_2
\]

6. Rearrange terms:

\[
0.5n_1 = 0.5n_2
\]

\[
n_1 = n_2
\]