Two containers of the same volume, one containing the gas A and the other containing the gas B, have the same number of molecules of each gas. The mass of the molecules A is twice the mass of the molecule B. The rms speed of A is also twice that of B. Calculate the pressure ratio of the gases A and B.

Answer

To solve the problem, we need to find the pressure ratio of gases A and B given the conditions about their molecular masses and root mean square (RMS) speeds. Here’s the step-by-step solution:

### Step 1: Understand the given information
- Let the mass of a molecule of gas B be \( m_B \).
- Then, the mass of a molecule of gas A is \( m_A = 2m_B \).
- The RMS speed of gas A is given as \( v_{rms,A} = 2v_{rms,B} \).

### Step 2: Write the formula for RMS speed
The formula for the RMS speed of a gas is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

### Step 3: Set up the equations for both gases
For gas A:
\[ v_{rms,A} = \sqrt{\frac{3RT}{m_A}} = \sqrt{\frac{3RT}{2m_B}} \]

For gas B:
\[ v_{rms,B} = \sqrt{\frac{3RT}{m_B}} \]

### Step 4: Relate the RMS speeds
According to the problem, \( v_{rms,A} = 2v_{rms,B} \):
\[ \sqrt{\frac{3RT}{2m_B}} = 2\sqrt{\frac{3RT}{m_B}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{3RT}{2m_B} = 4 \cdot \frac{3RT}{m_B} \]

### Step 6: Simplify the equation
Cancel \( 3RT \) from both sides (since \( R \) and \( T \) are the same for both gases):
\[ \frac{1}{2m_B} = \frac{4}{m_B} \]

### Step 7: Solve for the pressure ratio
From the ideal gas law, we know:
\[ PV = nRT \]
Since both containers have the same volume \( V \) and the same number of molecules, we can express the pressures as:
\[ P_A = \frac{n_A RT}{V} \quad \text{and} \quad P_B = \frac{n_B RT}{V} \]

Given that \( n_A = n_B \) and substituting \( n_A = \frac{N_A}{N_A + N_B} \) and \( n_B = \frac{N_B}{N_A + N_B} \), we can find the ratio of pressures:
\[ \frac{P_A}{P_B} = \frac{m_B v_{rms,B}^2}{m_A v_{rms,A}^2} \]

Substituting \( m_A = 2m_B \) and \( v_{rms,A} = 2v_{rms,B} \):
\[ \frac{P_A}{P_B} = \frac{m_B (v_{rms,B})^2}{2m_B (2v_{rms,B})^2} = \frac{1}{2 \cdot 4} = \frac{1}{8} \]

Thus, the pressure ratio is:
\[ P_A : P_B = 8 : 1 \]

### Final Answer
The pressure ratio of gases A and B is \( 8 : 1 \).

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