The total pressure of a mixture of 8g of oxygen and 14g of nitrogen contained in a 11.2L vessel at 0°C is.

Options

A. 0.5 atm
B. 1 atm
C. 1.5 atm
D. 2 atm

Answer

To find the total pressure of a mixture of 8g of oxygen and 14g of nitrogen contained in an 11.2L vessel at 0°C, we can follow these steps:

### Step 1: Calculate the number of moles of oxygen (O₂)
- The molar mass of oxygen (O₂) is 32 g/mol.
- Given mass of oxygen = 8 g.
- Number of moles of oxygen (n₁) = mass / molar mass = 8 g / 32 g/mol = 0.25 moles.

### Step 2: Calculate the number of moles of nitrogen (N₂)
- The molar mass of nitrogen (N₂) is 28 g/mol.
- Given mass of nitrogen = 14 g.
- Number of moles of nitrogen (n₂) = mass / molar mass = 14 g / 28 g/mol = 0.5 moles.

### Step 3: Calculate the total number of moles in the mixture
- Total number of moles (n) = n₁ + n₂ = 0.25 moles + 0.5 moles = 0.75 moles.

### Step 4: Convert the temperature from Celsius to Kelvin
- Given temperature = 0°C.
- Temperature in Kelvin (T) = 0 + 273 = 273 K.

### Step 5: Use the Ideal Gas Law to calculate the pressure
- The Ideal Gas Law is given by the equation: PV = nRT, where:
- P = pressure (in atm)
- V = volume (in L)
- n = number of moles
- R = universal gas constant = 0.0821 L·atm/(K·mol)
- T = temperature (in K)

Rearranging the equation to solve for P:
\[ P = \frac{nRT}{V} \]

### Step 6: Substitute the known values into the equation
- Substitute n = 0.75 moles, R = 0.0821 L·atm/(K·mol), T = 273 K, and V = 11.2 L into the equation:
\[ P = \frac{(0.75 \text{ moles}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (273 \text{ K})}{11.2 \text{ L}} \]

### Step 7: Calculate the pressure
- Performing the calculation:
\[ P = \frac{(0.75) \times (0.0821) \times (273)}{11.2} \]
\[ P = \frac{(0.75 \times 22.4143)}{11.2} \]
\[ P = \frac{16.810725}{11.2} \]
\[ P \approx 1.5 \text{ atm} \]

### Final Answer:
The total pressure of the mixture is approximately **1.5 atm**.

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