Solution:

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 \( P_1 > P_2 \).

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Solution:

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

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Solution:

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 \( V_2 > V_3 \).

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Solution:

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 \( 1.5V \).

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Solution:

To find the ratio of the masses of \( \text{N}_2 \) and \( \text{He} \) in vessels \( A \) and \( B \), we use the ideal gas law:

\[
PV = \frac{m}{M} RT
\]

where:
- \( m \) is the mass of the gas,
- \( M \) is the molar mass,
- \( P, V, R, T \) are pressure, volume, gas constant, and temperature, respectively.

For vessel \( A \) (containing \( \text{N}_2 \)):
\[
m_{\text{N}_2} = \frac{PVM_{\text{N}_2}}{RT}
\]

For vessel \( B \) (containing \( \text{He} \) at temperature \( 2T \)):
\[
m_{\text{He}} = \frac{PV M_{\text{He}}}{R \cdot 2T} = \frac{PVM_{\text{He}}}{2RT}
\]

Now, the ratio of the masses \( \frac{m_{\text{N}_2}}{m_{\text{He}}} \) is:

\[
\frac{m_{\text{N}_2}}{m_{\text{He}}} = \frac{\frac{PVM_{\text{N}_2}}{RT}}{\frac{PVM_{\text{He}}}{2RT}} = \frac{M_{\text{N}_2}}{M_{\text{He}}} \times 2
\]

Since \( M_{\text{N}_2} = 28 \) and \( M_{\text{He}} = 4 \):

\[
\frac{m_{\text{N}_2}}{m_{\text{He}}} = \frac{28}{4} \times 2 = 14
\]

Thus, the ratio of masses of \( \text{N}_2 \) to \( \text{He} \) is  14:1.

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Solution:

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.

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Solution:

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.

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Solution:

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.

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Solution:

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.

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Solution:

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

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