Two gases A and B, having the mole ratio of 3 : 5 in a container, exert a pressure of 8 atm. If A is removed, what would be the pressure due to B only, temperature remaining constant?

Options

A. 1 atm
B. 2 atm
C. 4 atm
D. 5 atm

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Given Information
We have two gases A and B in a container with a mole ratio of 3:5. The total pressure exerted by both gases is 8 atm.

### Step 2: Define the Moles of Gases
Let the number of moles of gas A be \(3x\) and the number of moles of gas B be \(5x\). This is based on the given mole ratio of 3:5.

### Step 3: Calculate the Total Moles
The total number of moles in the container can be calculated as:
\[ N_{total} = N_A + N_B = 3x + 5x = 8x \]

### Step 4: Calculate the Mole Fraction of Gas B
The mole fraction of gas B (\(X_B\)) can be calculated using the formula:
\[ X_B = \frac{N_B}{N_{total}} = \frac{5x}{8x} = \frac{5}{8} \]

### Step 5: Apply Dalton's Law of Partial Pressures
Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each gas. The partial pressure of gas B (\(P_B\)) can be calculated using the formula:
\[ P_B = X_B \cdot P_T \]
Where \(P_T\) is the total pressure (8 atm).

### Step 6: Calculate the Partial Pressure of Gas B
Substituting the values we have:
\[ P_B = \left(\frac{5}{8}\right) \cdot 8 \, \text{atm} = 5 \, \text{atm} \]

### Step 7: Conclusion
If gas A is removed, the pressure due to gas B only will be 5 atm.

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