One half mole each of nitrogen, oxygen and carbon dioxide are mixed in enclosure of volume 5 litres and temperature `27^(@) C`. Calculate the pressure exerted by the mixture. Given `R = 8.31 J mol^(-1) K^(-1)`.

Options

A. `7.48xx10^(5)Nm^(-2)`
B. `5xx10^(5)Nm^(2)`
C. `6xx10^(5)Nm^(2)`
D. `3xx10^(5)Nm^(-2)`

Explanation

To calculate the pressure exerted by the mixture of gases (nitrogen, oxygen, and carbon dioxide), we can use the ideal gas law and Dalton's law of partial pressures. Here’s a step-by-step solution:

### Step 1: Convert Temperature to Kelvin
The temperature is given as \(27^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
So,
\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 2: Identify Moles of Each Gas
We have:
- \(n_1 = n_2 = n_3 = \frac{1}{2} \, \text{moles}\) (for nitrogen, oxygen, and carbon dioxide respectively).

### Step 3: Calculate Total Moles
The total number of moles \(n_{total}\) in the mixture is:
\[ n_{total} = n_1 + n_2 + n_3 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \, \text{moles} \]

### Step 4: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \(P\) is the pressure,
- \(V\) is the volume (in cubic meters),
- \(n\) is the number of moles,
- \(R\) is the universal gas constant (\(8.31 \, J \, mol^{-1} \, K^{-1}\)),
- \(T\) is the temperature in Kelvin.

### Step 5: Substitute Values into the Ideal Gas Law
The volume \(V\) is given as \(5 \, \text{liters}\). We need to convert this to cubic meters:
\[ V = 5 \, \text{liters} = 5 \times 10^{-3} \, m^3 \]
Now substituting the values into the ideal gas law:
\[ P = \frac{nRT}{V} \]
Substituting \(n = \frac{3}{2}\), \(R = 8.31 \, J \, mol^{-1} \, K^{-1}\), and \(T = 300 \, K\):
\[ P = \frac{\left(\frac{3}{2}\right) \times 8.31 \times 300}{5 \times 10^{-3}} \]

### Step 6: Calculate the Pressure
Calculating the numerator:
\[ \frac{3}{2} \times 8.31 \times 300 = 1246.5 \, J \]
Now divide by the volume:
\[ P = \frac{1246.5}{5 \times 10^{-3}} = 249300 \, Pa = 249.3 \, kPa \]

### Final Result
The pressure exerted by the mixture is approximately:
\[ P \approx 249.3 \, kPa \]