To find the partial pressure of argon in the jar, we can use the Ideal Gas Law, which is stated as:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure of the gas,
- [tex]\( V \)[/tex] is the volume of the gas,
- [tex]\( n \)[/tex] is the number of moles of the gas,
- [tex]\( R \)[/tex] is the ideal gas constant,
- [tex]\( T \)[/tex] is the temperature in Kelvin.
Given:
- Volume [tex]\( V = 25 \)[/tex] liters
- Number of moles of argon [tex]\( n = 0.0104 \)[/tex] moles
- Temperature [tex]\( T = 273 \)[/tex] Kelvin
- Ideal gas constant [tex]\( R = 8.314 \, \frac{L \cdot kPa}{mol \cdot K} \)[/tex]
We need to solve for the pressure [tex]\( P \)[/tex]. Rearranging the ideal gas law equation to solve for [tex]\( P \)[/tex]:
[tex]\[ P = \frac{nRT}{V} \][/tex]
Substituting the given values into the equation:
[tex]\[ P = \frac{(0.0104 \, \text{mol}) \times (8.314 \, \frac{L \cdot kPa}{mol \cdot K}) \times (273 \, \text{K})}{25 \, \text{L}} \][/tex]
[tex]\[ P = \frac{23.6202432 \, L \cdot kPa}{25 \, L} \][/tex]
[tex]\[ P = 0.944204352 \, kPa \][/tex]
Expressing the answer to three significant figures:
[tex]\[ P = 0.944 \, kPa \][/tex]
Therefore, the partial pressure of argon in the jar is [tex]\( \boxed{0.944} \)[/tex] kilopascal.
