To determine the partial pressure of argon in a 25.0-liter jar at a temperature of 273 K containing 0.0104 moles of argon, we'll use the Ideal Gas Law, expressed as [tex]\( PV = nRT \)[/tex].
Here are the steps we need to follow:
1. Identify the given values:
- Volume ([tex]\(V\)[/tex]): 25.0 liters
- Moles of argon ([tex]\(n\)[/tex]): 0.0104 moles
- Temperature ([tex]\(T\)[/tex]): 273 Kelvin
- Ideal gas constant ([tex]\(R\)[/tex]): 8.314 [tex]\(\frac{L \cdot kPa}{mol \cdot K}\)[/tex]
2. Rearrange the Ideal Gas Law to solve for the partial pressure ([tex]\(P\)[/tex]):
The formula for the Ideal Gas Law is [tex]\( PV = nRT \)[/tex]. To find the partial pressure, [tex]\(P\)[/tex], we rearrange this equation:
[tex]\[
P = \frac{nRT}{V}
\][/tex]
3. Substitute the given values into the equation:
[tex]\[
P = \frac{(0.0104 \, \text{mol}) \times (8.314 \, \frac{\text{L} \cdot \text{kPa}}{\text{mol} \cdot \text{K}}) \times (273 \, \text{K})}{25.0 \, \text{L}}
\][/tex]
4. Calculate the partial pressure:
[tex]\[
P = \frac{(0.0104) \times (8.314) \times (273)}{25.0}
\][/tex]
After calculating, we find that:
[tex]\[
P \approx 0.944 \, \text{kPa}
\][/tex]
Therefore, the partial pressure of argon in the jar is 0.944 kilopascals.
