16 gram of oxygen, 14 gram of nitrogen and 11 gram of carbon dioxide are mixed in an enclosure of volume 5 L and temperature `27^(@)C. The pressure exerted by the mixture is

Options

A. `4xx10^(5)Nm^(-2)`
B. `5xx10^(5)Nm^(-2)`
C. `6xx10^(5)Nm^(-2)`
D. `9xx10^(5)Nm^(-2)`

Explanation

To find the pressure exerted by the mixture of gases, we can use the ideal gas law and Dalton's law of partial pressures. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of each gas.
Using the formula for moles:
\[ n = \frac{m}{M} \]
where \( m \) is the mass of the gas and \( M \) is the molar mass.

1. **For Oxygen (O2)**:
- Mass \( m = 16 \, \text{g} \)
- Molar mass \( M = 32 \, \text{g/mol} \)
\[ n_{O2} = \frac{16}{32} = 0.5 \, \text{mol} \]

2. **For Nitrogen (N2)**:
- Mass \( m = 14 \, \text{g} \)
- Molar mass \( M = 28 \, \text{g/mol} \)
\[ n_{N2} = \frac{14}{28} = 0.5 \, \text{mol} \]

3. **For Carbon Dioxide (CO2)**:
- Mass \( m = 11 \, \text{g} \)
- Molar mass \( M = 44 \, \text{g/mol} \)
\[ n_{CO2} = \frac{11}{44} = 0.25 \, \text{mol} \]

### Step 2: Calculate the total number of moles in the mixture.
\[ n_{total} = n_{O2} + n_{N2} + n_{CO2} = 0.5 + 0.5 + 0.25 = 1.25 \, \text{mol} \]

### Step 3: Use the ideal gas equation to find the pressure.
The ideal gas equation is given by:
\[ PV = nRT \]
Rearranging for pressure \( P \):
\[ P = \frac{nRT}{V} \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \) (or \( 8.314 \, \text{L kPa/(mol K)} \))
- \( T = 27^\circ C = 300 \, \text{K} \) (since \( T(K) = T(°C) + 273 \))
- \( V = 5 \, \text{L} \)

### Step 4: Substitute the values into the equation.
\[ P = \frac{(1.25 \, \text{mol}) \times (8.314 \, \text{L kPa/(mol K)}) \times (300 \, \text{K})}{5 \, \text{L}} \]

### Step 5: Calculate the pressure.
\[ P = \frac{(1.25) \times (8.314) \times (300)}{5} \]
Calculating the numerator:
\[ 1.25 \times 8.314 \times 300 = 9976.25 \, \text{kPa L} \]
Now divide by the volume:
\[ P = \frac{9976.25}{5} = 1995.25 \, \text{kPa} \]

### Step 6: Convert pressure to Newton per meter squared (if necessary).
Since \( 1 \, \text{kPa} = 1000 \, \text{N/m}^2 \):
\[ P = 1995.25 \times 1000 = 1995250 \, \text{N/m}^2 \]

### Final Answer
The pressure exerted by the mixture is approximately \( 2.0 \times 10^6 \, \text{N/m}^2 \).