To determine the final volume of the gas in the balloon, we can use the Combined Gas Law, which relates the pressure, volume, and temperature of a gas in two different states:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Where:
- [tex]\( P_1 \)[/tex] and [tex]\( P_2 \)[/tex] are the initial and final pressures, respectively.
- [tex]\( V_1 \)[/tex] and [tex]\( V_2 \)[/tex] are the initial and final volumes, respectively.
- [tex]\( T_1 \)[/tex] and [tex]\( T_2 \)[/tex] are the initial and final temperatures, respectively.
### Given Values:
- Initial volume ([tex]\( V_1 \)[/tex]): 25.8 L
- Initial temperature ([tex]\( T_1 \)[/tex]): 353 K
- Initial pressure ([tex]\( P_1 \)[/tex]): 2575 mm Hg
- Final pressure ([tex]\( P_2 \)[/tex]): 1.35 atm
- Final temperature ([tex]\( T_2 \)[/tex]): 253 K
### Step-by-Step Solution:
1. Convert Initial Pressure to the Same Units as Final Pressure (atm):
- We know that [tex]\( 1 \text{ atm} = 760 \text{ mm Hg} \)[/tex].
- Hence, [tex]\( P_1 \)[/tex] in atm:
[tex]\[
P_1 = \frac{2575 \text{ mm Hg}}{760 \text{ mm Hg/atm}} \approx 3.388 \text{ atm}
\][/tex]
2. Apply the Combined Gas Law to Solve for the Final Volume ([tex]\( V_2 \)[/tex]):
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/tex]
Re-arrange for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}
\][/tex]
3. Substitute the Known Values:
[tex]\[
V_2 = \frac{(3.388 \text{ atm}) (25.8 \text{ L}) (253 \text{ K})}{(353 \text{ K}) (1.35 \text{ atm})}
\][/tex]
Simplify and calculate [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 \approx 46.408 \text{ L}
\][/tex]
Therefore, the volume of gas the balloon will contain at 1.35 atm and 253 K is approximately 46.408 L. Hence, the closest option to this result is:
45.8 L
