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

Options

A. 0.492atm
B. 49.2atm
C. 4.52atm
D. 0.0492atm

Answer

To calculate the total pressure in a 10.0 L cylinder containing 0.4 g of helium, 1.6 g of oxygen, and 1.4 g of nitrogen at 27°C, we will use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

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

### Step 1: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Given that the temperature is 27°C:

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

### Step 2: Calculate the number of moles of each gas
We will calculate the number of moles for each gas 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

\[ n_{O2} = \frac{1.6 \, g}{32 \, g/mol} = 0.05 \, mol \]

**For Nitrogen (N2):**
- Mass = 1.4 g
- Molar mass = 28 g/mol

\[ n_{N2} = \frac{1.4 \, g}{28 \, g/mol} = 0.05 \, mol \]

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

\[ n_{total} = n_{He} + n_{O2} + n_{N2} \]

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

### Step 4: Use the ideal gas law to calculate the pressure
Now we can use the ideal gas law to find the pressure:

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

Substituting the values we have:

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

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

Calculating the numerator:

\[ 0.2 \times 0.0821 \times 300.15 = 4.926 \]

Now divide by the volume:

\[ P = \frac{4.926}{10.0} = 0.4926 \, atm \]

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

\[ P \approx 0.493 \, atm \]