Calculate the total pressure (in atm) in a 10.0 L cylinder which contains 0.4 g helium, 1.6 g oxygen and 1.4 g nitrogen at 27 °C.

Explanation

To calculate the total pressure in the cylinder, we will use the Ideal Gas Law, which is given by the formula:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature in Celsius is given as 27 °C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Calculating:

\[ T = 27 + 273.15 = 300.15 \, K \]

### Step 2: Calculate the number of moles of each gas
We will calculate the number of moles for helium (He), oxygen (O₂), and nitrogen (N₂) using the formula:

\[ n = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- **For Helium (He)**:
- Mass = 0.4 g
- Molar mass = 4 g/mol
\[ n_{He} = \frac{0.4 \, g}{4 \, g/mol} = 0.1 \, mol \]

- **For Oxygen (O₂)**:
- Mass = 1.6 g
- Molar mass = 32 g/mol
\[ n_{O_2} = \frac{1.6 \, g}{32 \, g/mol} = 0.05 \, mol \]

- **For Nitrogen (N₂)**:
- Mass = 1.4 g
- Molar mass = 28 g/mol
\[ n_{N_2} = \frac{1.4 \, g}{28 \, g/mol} = 0.05 \, mol \]

### Step 3: Calculate the total number of moles
Now, we sum the moles of all gases:

\[ n_{total} = n_{He} + n_{O_2} + n_{N_2} = 0.1 + 0.05 + 0.05 = 0.2 \, mol \]

### Step 4: Calculate the total pressure using the Ideal Gas Law
Now we can substitute the values into the Ideal Gas Law rearranged to solve for pressure:

\[ P = \frac{nRT}{V} \]

Where:
- \( n = 0.2 \, mol \)
- \( R = 0.0821 \, L·atm/(K·mol) \)
- \( T = 300.15 \, K \)
- \( V = 10.0 \, L \)

Calculating:

\[ P = \frac{(0.2 \, mol) \times (0.0821 \, L·atm/(K·mol)) \times (300.15 \, K)}{10.0 \, L} \]

Calculating the numerator:

\[ P = \frac{(0.2) \times (0.0821) \times (300.15)}{10.0} \]
\[ P = \frac{4.9263}{10.0} = 0.49263 \, atm \]

### Final Answer
The total pressure in the cylinder is approximately:

\[ P \approx 0.493 \, atm \]