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.

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

To analyze the \( V \) versus \( T \) curves for an ideal gas at constant pressures \( P_1 \) and \( P_2 \):

According to Charles's Law:

\[
\frac{V}{T} = \frac{nR}{P}
\]

This implies:

1. At constant pressure, the volume \( V \) is directly proportional to the temperature \( T \), resulting in a straight line.

2. For a given temperature, the volume \( V \) will be greater at a lower pressure \( P \), because \( V \propto \frac{1}{P} \) for a fixed amount of gas.

In the graph:
- The line with a steeper slope corresponds to a lower pressure (since a lower \( P \) results in a larger \( V \) for the same \( T \)).

Since \( P_2 \) has a steeper slope than \( P_1 \), we conclude that:

\[
P_1 > P_2
\]

Therefore, the correct answer is \( P_1 > P_2 \).

Given:

- The increase in temperature \(\Delta T = 2^\circ \text{C} = 2 \text{ K}\)
- The increase in pressure \(\frac{\Delta P}{P} = 2\% = 0.02\)

Using the relation for a gas at constant volume, \(\frac{\Delta P}{P} = \frac{\Delta T}{T}\):

\[
0.02 = \frac{2}{T}
\]

Solving for \(T\):

\[
T = \frac{2}{0.02} = 100 \text{ K}
\]

Answer: \(T = 100 \, \text{K}\)

Since the temperature is constant, we can use Boyle's Law, which states that \( P_1 V_1 = P_2 V_2 \).

Given:
- Initial pressure, \( P_1 = 2 \) atm
- Initial volume, \( V_1 = 3V \)
- Final pressure, \( P_2 = 2 \times P_1 = 4 \) atm

Using Boyle's Law:

\[
P_1 V_1 = P_2 V_2
\]
\[
2 \times 3V = 4 \times V_2
\]
\[
6V = 4V_2
\]
\[
V_2 = \frac{6V}{4} = 1.5V
\]

So, the final volume \( V_2 \) is \( 1.5V \).

At constant volume and temperature, the ideal gas law is:

\[
PV = nRT
\]

Since \( n = \frac{\text{mass}}{\text{molar mass}} \), we can rewrite the equation as:

\[
P \propto \frac{\text{mass}}{M}
\]

Thus, for a given volume and temperature, the pressure \( P \) varies directly with the mass of the gas.