Given that density \(\rho\) is constant, we use the ideal gas law:

\[
\frac{P}{T} = \text{constant (since } \rho \text{ is constant)}
\]

If the temperature \(T\) is doubled, then:

\[
\frac{P_2}{T_2} = \frac{P_1}{T_1} \Rightarrow \frac{P_2}{2T_1} = \frac{P_1}{T_1}
\]

Thus, \(P_2 = 2P_1\), meaning the pressure also doubles, resulting in a 100% increase.

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

Since both gases \( A \) and \( B \) have the same temperature \( T \), pressure \( P \), and volume \( V \), each gas has the same number of moles, \( n \), by the ideal gas law:

\[
PV = nRT \Rightarrow n = \frac{PV}{RT}
\]

When gases \( A \) and \( B \) are mixed, the total number of moles becomes \( 2n \). Since temperature \( T \) and volume \( V \) remain the same, the total pressure \( P_{\text{mixture}} \) will be:

\[
P_{\text{mixture}} V = (2n)RT
\]

\[
P_{\text{mixture}} = 2P
\]

Answer: The pressure of the mixture is \( 2P \).

In the gas equation \( PV = nRT \), the universal gas constant \( R \) has a fixed value but its numerical value depends on the units of \( P \), \( V \), and \( T \).

Answer: The value of the universal gas constant depends only on the units of measurement.

The expression \(\frac{PV}{kT}\) can be analyzed using the ideal gas law:

\[
PV = NkT
\]

where:
- \( P \) is pressure,
- \( V \) is volume,
- \( N \) is the number of molecules,
- \( k \) is Boltzmann's constant, and
- \( T \) is temperature.

Rearranging, we get:

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

Answer: This quantity represents the "number of molecules in the gas".

Given:

- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- Final temperature, \( T_2 = 87^\circ \text{C} = 360 \, \text{K} \)
- Initial pressure, \( P_1 = 2 \, \text{atm} \)

Using the pressure-temperature relationship for a gas at constant volume:

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

Solving for \( P_2 \):

\[
P_2 = P_1 \times \frac{T_2}{T_1} = 2 \times \frac{360}{300} = 2 \times 1.2 = 2.4 \, \text{atm}
\]

Answer: \( P_2 = 2.4 \, \text{atm} \)

Given:

- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- We want to expel half the mass of air, meaning the final mass, \( m_2 = \frac{m_1}{2} \).

Using the ideal gas law, \( PV = nRT \), and since pressure and volume are constant, \( \frac{m}{T} = \text{constant} \).

Thus,

\[
\frac{m_1}{T_1} = \frac{m_2}{T_2}
\]

Substitute \( m_2 = \frac{m_1}{2} \):

\[
\frac{m_1}{300} = \frac{\frac{m_1}{2}}{T_2}
\]

Solving for \( T_2 \):

\[
T_2 = 2 \times 300 = 600 \, \text{K}
\]

Converting back to Celsius:

\[
T_2 = 600 - 273 = 327^\circ \text{C}
\]

Answer: \( T_2 = 327^\circ \text{C} \)

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