Calculate the total pressure in a mixture of `8 g` of oxygen and `4 g` hydrogen confined in a vessel of `1 dm^(3)` at `27^(@)C`. `(R = 0.083 "bar" dm^(3) K^(-1) mol^(-1))`

Explanation

To calculate the total pressure in a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm³ at 27°C, we will follow these steps:

### Step 1: Calculate the number of moles of oxygen (O₂)
The molecular weight of oxygen (O₂) is 32 g/mol.
Using the formula for moles:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]
For oxygen:
\[ n_{O_2} = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = 0.25 \, \text{mol} \]

### Step 2: Calculate the number of moles of hydrogen (H₂)
The molecular weight of hydrogen (H₂) is 2 g/mol.
Using the same formula for moles:
\[ n_{H_2} = \frac{4 \, \text{g}}{2 \, \text{g/mol}} = 2 \, \text{mol} \]

### Step 3: Convert the temperature from Celsius to Kelvin
The temperature given is 27°C. To convert to Kelvin:
\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 4: Use the ideal gas equation to calculate the partial pressure of oxygen
The ideal gas equation is:
\[ PV = nRT \]
Rearranging for pressure (P):
\[ P = \frac{nRT}{V} \]
Substituting the values for oxygen:
- \(n_{O_2} = 0.25 \, \text{mol}\)
- \(R = 0.083 \, \text{bar dm}^3 \, \text{K}^{-1} \, \text{mol}^{-1}\)
- \(T = 300 \, \text{K}\)
- \(V = 1 \, \text{dm}^3\)

Calculating:
\[ P_{O_2} = \frac{0.25 \times 0.083 \times 300}{1} = 6.225 \, \text{bar} \]

### Step 5: Use the ideal gas equation to calculate the partial pressure of hydrogen
Using the same ideal gas equation for hydrogen:
Substituting the values for hydrogen:
- \(n_{H_2} = 2 \, \text{mol}\)

Calculating:
\[ P_{H_2} = \frac{2 \times 0.083 \times 300}{1} = 49.8 \, \text{bar} \]

### Step 6: Calculate the total pressure
The total pressure (P_total) is the sum of the partial pressures:
\[ P_{\text{total}} = P_{O_2} + P_{H_2} = 6.225 + 49.8 = 56.025 \, \text{bar} \]

### Final Answer
The total pressure in the mixture is **56.025 bar**.

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