To determine the final volume of the gas when its pressure and temperature change, we can use the combined gas law:
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/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,
- [tex]\( V_2 \)[/tex] is the final volume,
- [tex]\( T_2 \)[/tex] is the final temperature (in Kelvin).
Given:
[tex]\[
P_1 = 1.5 \, \text{atm} \\
V_1 = 3.0 \, \text{L} \\
T_1 = 20.0^{\circ} \text{C} = 20.0 + 273.15 = 293.15 \, \text{K} \\
P_2 = 2.5 \, \text{atm} \\
T_2 = 30.0^{\circ} \text{C} = 30.0 + 273.15 = 303.15 \, \text{K}
\][/tex]
We need to solve for the final volume, [tex]\( V_2 \)[/tex]. Rearrange the combined gas law to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
\][/tex]
Substitute in the known values:
[tex]\[
V_2 = \frac{(1.5 \, \text{atm}) (3.0 \, \text{L}) (303.15 \, \text{K})}{(2.5 \, \text{atm}) (293.15 \, \text{K})}
\][/tex]
[tex]\[
V_2 = \frac{1.5 \times 3.0 \times 303.15}{2.5 \times 293.15}
\][/tex]
[tex]\[
V_2 \approx 1.861
\][/tex]
Therefore, the final volume [tex]\( V_2 \)[/tex] that the gas will occupy under the new conditions is approximately 1.861 L.
