To determine the volume of dry Xenon gas at standard temperature and pressure (STP), we will utilize the combined gas law. This law relates the initial and final conditions of a gas sample in terms of pressure (P), volume (V), and temperature (T):
[tex]\[
\frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}
\][/tex]
where:
- [tex]\(P_1\)[/tex] is the initial pressure
- [tex]\(V_1\)[/tex] is the initial volume
- [tex]\(T_1\)[/tex] is the initial temperature in Kelvin
- [tex]\(P_2\)[/tex] is the final pressure (at STP)
- [tex]\(V_2\)[/tex] is the final volume (at STP, which we need to determine)
- [tex]\(T_2\)[/tex] is the final temperature (at STP) in Kelvin
Given data:
- [tex]\(P_1 = 205 \, \text{kPa}\)[/tex]
- [tex]\(V_1 = 6.12 \, \text{L}\)[/tex]
- Initial temperature [tex]\(T_1 = 80 \, ^\circ \text{C}\)[/tex]
Converting the initial temperature to Kelvin:
[tex]\[
T_1 = 80 + 273.15 = 353.15 \, \text{K}
\][/tex]
Standard temperature and pressure (STP) conditions:
- [tex]\(P_2 = 101.325 \, \text{kPa}\)[/tex]
- Standard temperature [tex]\(T_2 = 0 \, ^\circ \text{C}\)[/tex] which is:
[tex]\[
T_2 = 0 + 273.15 = 273.15 \, \text{K}
\][/tex]
Rewriting the combined gas law to solve for [tex]\(V_2\)[/tex]:
[tex]\[
V_2 = \frac{P_1 \times V_1 \times T_2 }{P_2 \times T_1}
\][/tex]
Substitute the known values into the equation:
[tex]\[
V_2 = \frac{205 \, \text{kPa} \times 6.12 \, \text{L} \times 273.15 \, \text{K}}{101.325 \, \text{kPa} \times 353.15 \, \text{K}}
\][/tex]
Carry out the multiplication and division:
[tex]\[
V_2 = \frac{205 \times 6.12 \times 273.15}{101.325 \times 353.15}
\][/tex]
[tex]\[
V_2 = \frac{3437.226}{35769.73875}
\][/tex]
[tex]\[
V_2 \approx 2.87 \, \text{L}
\][/tex]
Therefore, the volume of the dry Xenon gas at STP is approximately [tex]\(2.87 \, \text{L}\)[/tex].
