A `4.0 dm^(3)` flask containing `N_(2) at 4` bar was connected to a `6.0 dm^(3)` flask containing helium at `6` bar , and the gases were allowed to mix isothermally. The total pressure of the resulting mixture will be

Options

A. 10.0 bar
B. 5.2 bar
C. 1.6 bar
D. 5.0 bar

Explanation

To find the total pressure of the resulting mixture when the gases are allowed to mix isothermally, we can use Dalton's Law of Partial Pressures and the ideal gas law.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- For Nitrogen (N₂):
- Volume (V₁) = 4.0 dm³
- Pressure (P₁) = 4.0 bar
- For Helium (He):
- Volume (V₂) = 6.0 dm³
- Pressure (P₂) = 6.0 bar

2. **Calculate the Moles of Each Gas:**
Using the ideal gas equation \( PV = nRT \), we can express the number of moles (n) in terms of pressure and volume. Since the temperature (T) and the gas constant (R) are constant, we can ignore them for this calculation.

- Moles of N₂ (n₁):
\[ n₁ = \frac{P₁ \cdot V₁}{RT} \quad \text{(we can ignore RT for comparison)} \]
\[ n₁ = P₁ \cdot V₁ = 4 \, \text{bar} \cdot 4 \, \text{dm}^3 = 16 \, \text{bar dm}^3 \]

- Moles of He (n₂):
\[ n₂ = \frac{P₂ \cdot V₂}{RT} \quad \text{(we can ignore RT for comparison)} \]
\[ n₂ = P₂ \cdot V₂ = 6 \, \text{bar} \cdot 6 \, \text{dm}^3 = 36 \, \text{bar dm}^3 \]

3. **Calculate the Total Moles:**
The total moles of gas after mixing is:
\[ n_{total} = n₁ + n₂ = 16 + 36 = 52 \, \text{bar dm}^3 \]

4. **Calculate the Total Volume:**
The total volume (V_total) after mixing is:
\[ V_{total} = V₁ + V₂ = 4 \, \text{dm}^3 + 6 \, \text{dm}^3 = 10 \, \text{dm}^3 \]

5. **Calculate the Total Pressure:**
Using the formula \( P_{total} = \frac{n_{total}}{V_{total}} \):
\[ P_{total} = \frac{52 \, \text{bar dm}^3}{10 \, \text{dm}^3} = 5.2 \, \text{bar} \]

### Final Answer:
The total pressure of the resulting mixture will be **5.2 bar**.