Using the ideal gas law \( PV = nRT \), we find the moles \( n \) in each vessel:

1. For vessel \( A \) with \( O_2 \):
\[
n_{\text{O}_2} = \frac{PV}{RT}
\]

2. For vessel \( B \) with \( H_2 \) at \( 2P \):
\[
n_{\text{H}_2} = \frac{2PV}{RT} = 2 \times \frac{PV}{RT}
\]

Now, the mass \( m = n \times \text{molar mass} \):

- Mass of \( O_2 \) in \( A = n_{\text{O}_2} \times 32 = \frac{PV}{RT} \times 32 \)
- Mass of \( H_2 \) in \( B = n_{\text{H}_2} \times 2 = 2 \times \frac{PV}{RT} \times 2 \)

So, the mass ratio is:
\[
\frac{\text{mass of } O_2}{\text{mass of } H_2} = \frac{\frac{PV}{RT} \times 32}{2 \times \frac{PV}{RT} \times 2} = \frac{32}{4} = 8:1
\]

For one mole of an ideal gas, we use the ideal gas law:

\[
PV = RT
\]

Given \( P = \frac{P_0}{1 + \left( \frac{V_0}{V} \right)^2} \), find \( T \) at \( V = V_0 \) and \( V = 2V_0 \).

1. When \( V = V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{V_0} \right)^2} = \frac{P_0}{1 + 1} = \frac{P_0}{2}
\]
\[
T_1 = \frac{PV}{R} = \frac{\frac{P_0}{2} \cdot V_0}{R} = \frac{P_0 V_0}{2R}
\]

2. When \( V = 2V_0 \):
\[
P = \frac{P_0}{1 + \left( \frac{V_0}{2V_0} \right)^2} = \frac{P_0}{1 + \frac{1}{4}} = \frac{P_0}{\frac{5}{4}} = \frac{4P_0}{5}
\]
\[
T_2 = \frac{PV}{R} = \frac{\frac{4P_0}{5} \cdot 2V_0}{R} = \frac{8P_0 V_0}{5R}
\]

3. Change in temperature:
\[
\Delta T = T_2 - T_1 = \frac{8P_0 V_0}{5R} - \frac{P_0 V_0}{2R} = \frac{11P_0 V_0}{10R}
\]

Given:

\[
P = \frac{A}{1 + \left( \frac{B}{V} \right)^2}
\]

Using the ideal gas equation, \( PV = nRT \), for initial and final states, we can express the temperature change.

 Step 1: Initial State (at \( V = B \))
\[
P_1 = \frac{A}{1 + \left( \frac{B}{B} \right)^2} = \frac{A}{2}
\]
\[
T_1 = \frac{P_1 V}{R} = \frac{\left(\frac{A}{2}\right) B}{R} = \frac{AB}{2R}
\]

Step 2: Final State (at \( V = 2B \))
\[
P_2 = \frac{A}{1 + \left( \frac{B}{2B} \right)^2} = \frac{A}{1 + \frac{1}{4}} = \frac{A}{\frac{5}{4}} = \frac{4A}{5}
\]
\[
T_2 = \frac{P_2 V}{R} = \frac{\left(\frac{4A}{5}\right) (2B)}{R} = \frac{8AB}{5R}
\]

 Step 3: Temperature Change
\[
\Delta T = T_2 - T_1 = \frac{8AB}{5R} - \frac{AB}{2R} = \frac{16AB - 5AB}{10R} = \frac{11AB}{10R}
\]

Answer: \(\frac{11AB}{10R}\)

For an ideal gas, at constant volume, the pressure \( P \) is directly proportional to the temperature \( T \) (Gay-Lussac's Law):

\[
P \propto T
\]

In the graph, each line represents a pressure-temperature relationship at different volumes. A steeper slope indicates a smaller volume (since \( P = \frac{nRT}{V} \) and a higher \( V \) gives a less steep line).

From the figure:
1. Lines \( 1 \) and \( 2 \) have the same slope, indicating \( V_1 = V_2 \).
2. Lines \( 3 \) and \( 4 \) also have the same slope, indicating \( V_3 = V_4 \).
3. The slope of lines \( 3 \) and \( 4 \) is steeper than that of lines \( 1 \) and \( 2 \), meaning \( V_2 > V_3 \).

Thus, the correct answer is \( V_1 = V_2 \), \( V_3 = V_4 \), and \( V_2 > V_3 \).

To solve this, we can use Gay-Lussac's Law, which states:

\[
\frac{P_1}{T_1} = \frac{P_2}{T_2}
\]

Given:
- Initial pressure \( P_1 = 870 \, \text{mm Hg} \),
- Final pressure \( P_2 = 1800 \, \text{mm Hg} \),
- Initial temperature \( T_1 = 17^\circ \text{C} = 17 + 273 = 290 \, \text{K} \).

Rearrange to find \( T_2 \):

\[
T_2 = \frac{P_2 \times T_1}{P_1}
\]

Substitute the values:

\[
T_2 = \frac{1800 \times 290}{870}
\]

Calculating this:

\[
T_2 = 600 \, \text{K}
\]

So, the temperature at which the pressure becomes 1800 mm of Hg is  600 K.

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 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 \( 4.06 \times 10^{24} \).

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 \( O_2 \)**.