To solve the problem of calculating the pressure of \(10^{22}\) molecules of sulfur dioxide (\(SO_2\)) in a 2.5 L vessel at a temperature of \(27^{\circ}C\), we will follow these steps:
### Step 1: Calculate the number of moles of \(SO_2\)
To find the number of moles, we use the formula:
\[
\text{Number of moles} (n) = \frac{\text{Number of molecules}}{\text{Avogadro's number}}
\]
Given:
- Number of molecules = \(10^{22}\)
- Avogadro's number = \(6.022 \times 10^{23} \, \text{molecules/mol}\)
Substituting the values:
\[
n = \frac{10^{22}}{6.022 \times 10^{23}} \approx 1.66 \times 10^{-2} \, \text{moles}
\]
### Step 2: 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:
- Temperature = \(27^{\circ}C\)
Calculating:
\[
T = 27 + 273.15 = 300.15 \, K \approx 300 \, K
\]
### Step 3: Use the Ideal Gas Law to calculate pressure
The Ideal Gas Law is given by:
\[
PV = nRT
\]
Where:
- \(P\) = pressure (in atm)
- \(V\) = volume (in liters)
- \(n\) = number of moles
- \(R\) = ideal gas constant (\(0.083 \, \text{L} \cdot \text{bar} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}\))
- \(T\) = temperature (in Kelvin)
We need to rearrange the equation to solve for pressure \(P\):
\[
P = \frac{nRT}{V}
\]
Substituting the known values:
\[
P = \frac{(1.66 \times 10^{-2} \, \text{mol}) \times (0.083 \, \text{L} \cdot \text{bar} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}) \times (300 \, K)}{2.5 \, \text{L}}
\]
Calculating the pressure:
\[
P = \frac{(1.66 \times 10^{-2}) \times (0.083) \times (300)}{2.5}
\]
\[
P \approx \frac{4.1318 \times 10^{-2}}{2.5} \approx 0.01652 \, \text{bar}
\]
### Step 4: Convert pressure from bar to atm (if needed)
To convert from bar to atm, we use the conversion factor \(1 \, \text{bar} \approx 0.9869 \, \text{atm}\):
\[
P \approx 0.01652 \, \text{bar} \times 0.9869 \, \text{atm/bar} \approx 0.0163 \, \text{atm}
\]
### Final Answer
The pressure of \(10^{22}\) molecules of sulfur dioxide in a 2.5 L vessel at \(27^{\circ}C\) is approximately:
\[
P \approx 0.0163 \, \text{atm}
\]
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