To determine the volume occupied by 11.5 grams of argon gas at a pressure of 1.29 atm and a temperature of 432 K, we should use the ideal gas law. The ideal gas law is given by the equation:
[tex]\[ PV = nRT \][/tex]
where:
- [tex]\( P \)[/tex] is the pressure,
- [tex]\( V \)[/tex] is the volume,
- [tex]\( n \)[/tex] is the number of moles,
- [tex]\( R \)[/tex] is the ideal gas constant,
- [tex]\( T \)[/tex] is the temperature.
Here’s how to solve this step-by-step:
1. Calculate the number of moles ([tex]\( n \)[/tex]):
We need to determine the amount of argon gas in moles. The molar mass of argon ([tex]\( \text{Ar} \)[/tex]) is approximately 39.948 g/mol.
The number of moles can be calculated using the formula:
[tex]\[
n = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
For 11.5 grams of argon:
[tex]\[
n = \frac{11.5 \, \text{g}}{39.948 \, \text{g/mol}} \approx 0.2879 \, \text{moles}
\][/tex]
2. Rearrange the ideal gas law to solve for the volume ([tex]\( V \)[/tex]):
[tex]\[
V = \frac{nRT}{P}
\][/tex]
3. Plug in the known values:
- [tex]\( n \approx 0.2879 \, \text{moles} \)[/tex]
- [tex]\( R = 0.0821 \, \text{L·atm/(mol·K)} \)[/tex]
- [tex]\( T = 432 \, \text{K} \)[/tex]
- [tex]\( P = 1.29 \, \text{atm} \)[/tex]
Now, substitute these values into the equation:
[tex]\[
V = \frac{0.2879 \, \text{moles} \times 0.0821 \, \text{L·atm/(mol·K)} \times 432 \, \text{K}}{1.29 \, \text{atm}}
\][/tex]
4. Calculate the volume:
[tex]\[
V \approx 7.9148 \, \text{L}
\][/tex]
Thus, the volume occupied by 11.5 grams of argon gas at a pressure of 1.29 atm and a temperature of 432 K is approximately 7.9148 liters.
