A vessel contains `0.5g` of hydrogen and `0.8g` of oxygen. The volume of vessel is `(5)/(28)m^(3)` while its temperature is maintanied at `300K`. Find the pressure of the mixture. (Assuming no reaction takes place between hydrogen and oxygen) `(M_(H_(2)) =2 , M_(O_(2)) = 32 & R = (25)/(3) J//mol-K)`

Explanation

To find the pressure of the mixture of hydrogen and oxygen in the vessel, we can follow these steps:

### Step 1: Calculate the number of moles of hydrogen (H₂)
The number of moles can be calculated using the formula:
\[ n = \frac{m}{M} \]
where \(m\) is the mass of the gas and \(M\) is the molar mass.

For hydrogen:
- Mass (\(m_{H_2}\)) = 0.5 g = 0.0005 kg (convert grams to kilograms)
- Molar mass (\(M_{H_2}\)) = 2 g/mol = 0.002 kg/mol

Calculating the number of moles of hydrogen:
\[ n_{H_2} = \frac{0.0005 \text{ kg}}{0.002 \text{ kg/mol}} = 0.25 \text{ mol} \]

### Step 2: Calculate the number of moles of oxygen (O₂)
Using the same formula for oxygen:
- Mass (\(m_{O_2}\)) = 0.8 g = 0.0008 kg
- Molar mass (\(M_{O_2}\)) = 32 g/mol = 0.032 kg/mol

Calculating the number of moles of oxygen:
\[ n_{O_2} = \frac{0.0008 \text{ kg}}{0.032 \text{ kg/mol}} = 0.025 \text{ mol} \]

### Step 3: Calculate the total number of moles in the mixture
The total number of moles (\(n_{total}\)) is the sum of the moles of hydrogen and oxygen:
\[ n_{total} = n_{H_2} + n_{O_2} = 0.25 \text{ mol} + 0.025 \text{ mol} = 0.275 \text{ mol} \]

### Step 4: Use the Ideal Gas Law to find the pressure
The Ideal Gas Law is given by:
\[ PV = nRT \]
We need to solve for pressure (\(P\)):
\[ P = \frac{nRT}{V} \]
Where:
- \(R = \frac{25}{3} \text{ J/(mol·K)}\)
- \(T = 300 \text{ K}\)
- \(V = \frac{5}{28} \text{ m}^3\)

Substituting the values into the equation:
\[ P = \frac{0.275 \text{ mol} \times \frac{25}{3} \text{ J/(mol·K)} \times 300 \text{ K}}{\frac{5}{28} \text{ m}^3} \]

### Step 5: Calculate the pressure
Calculating the numerator:
\[ 0.275 \times \frac{25}{3} \times 300 = 0.275 \times 2500 = 687.5 \text{ J} \]

Now, calculating the pressure:
\[ P = \frac{687.5 \text{ J}}{\frac{5}{28} \text{ m}^3} = 687.5 \times \frac{28}{5} = 687.5 \times 5.6 = 3848 \text{ Pa} \approx 3850 \text{ Pa} \]

### Final Answer
The pressure of the mixture is approximately \(3850 \text{ N/m}^2\).

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