A closed vessel contains a mixture of two diatomic gases `A` and `B`, Molar mass of `A` is 16times than of `B` and mass of gas `A` contained in the vessel is 2 times that of `B`. If the pressure exerted by `B` is `x` times of that exerted by `A`, then find the value of `x`.

Answer

To solve the problem, we need to find the relationship between the pressures exerted by the two gases \( A \) and \( B \) in a closed vessel. Here are the steps to derive the value of \( x \):

### Step 1: Identify the given information
- Molar mass of gas \( A \) is \( 16 \times \) that of gas \( B \).
- Mass of gas \( A \) is \( 2 \times \) that of gas \( B \).

Let:
- Molar mass of gas \( B \) = \( M_B \)
- Molar mass of gas \( A \) = \( 16M_B \)
- Mass of gas \( B \) = \( m_B \)
- Mass of gas \( A \) = \( 2m_B \)

### Step 2: Calculate the number of moles of each gas
Using the formula for the number of moles:
\[ n = \frac{m}{M} \]
- Number of moles of gas \( A \):
\[ n_A = \frac{m_A}{M_A} = \frac{2m_B}{16M_B} = \frac{m_B}{8M_B} \]

- Number of moles of gas \( B \):
\[ n_B = \frac{m_B}{M_B} \]

### Step 3: Write the relationship between pressures
According to the ideal gas law, the pressure exerted by a gas is directly proportional to the number of moles when volume and temperature are constant:
\[ P_A \propto n_A \quad \text{and} \quad P_B \propto n_B \]

Thus, we can express the pressures as:
\[ \frac{P_B}{P_A} = \frac{n_B}{n_A} \]

### Step 4: Substitute the values of \( n_A \) and \( n_B \)
Substituting the expressions for \( n_A \) and \( n_B \):
\[ \frac{P_B}{P_A} = \frac{n_B}{n_A} = \frac{\frac{m_B}{M_B}}{\frac{m_B}{8M_B}} = \frac{8M_B}{1} = 8 \]

### Step 5: Relate the pressures to find \( x \)
From the equation above, we have:
\[ P_B = 8P_A \]
This implies that the pressure exerted by gas \( B \) is \( 8 \) times that exerted by gas \( A \):
\[ x = 8 \]

### Final Answer
Thus, the value of \( x \) is \( 8 \).

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