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 will use Dalton's Law of Partial Pressures and the concept of mole fractions.

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

### Step 2: Define the moles of gases A and B
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 mole ratio provided (3:5).

### Step 3: Calculate the total number of moles
The total number of moles in the container is the sum of the moles of A and B:
\[ \text{Total moles} = 3x + 5x = 8x \]

### Step 4: Calculate the mole fraction of gas B
The mole fraction of gas B (\(X_B\)) is given by the formula:
\[ X_B = \frac{\text{moles of B}}{\text{total moles}} = \frac{5x}{8x} = \frac{5}{8} \]

### Step 5: Use Dalton's Law to find the partial pressure of gas B
According to Dalton's Law of Partial Pressures, the partial pressure of gas B (\(P_B\)) can be calculated as:
\[ P_B = X_B \times P_{\text{total}} \]
Substituting the values we have:
\[ P_B = \left(\frac{5}{8}\right) \times 8 \text{ atm} \]

### Step 6: Simplify to find the pressure of gas B
Calculating the above expression:
\[ P_B = \frac{5 \times 8}{8} = 5 \text{ atm} \]

### Step 7: Conclusion
Therefore, the pressure due to gas B only, after gas A is removed and with temperature remaining constant, is **5 atm**.