To solve this problem, we use the Combined Gas Law, which relates pressure (P), volume (V), and temperature (T) of a gas before and after a change.
The Combined Gas Law is given by:
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/tex]
We need to find the final pressure \( P_2 \). Before that, let's list the given values:
- Initial volume \( V_1 \) = 2.19 L
- Initial temperature \( T_1 \) = 47°C = 47 + 273.15 = 320.15 K
- Initial pressure \( P_1 \) = 805 mmHg
- Final volume \( V_2 \) = 9.88 L
- Final temperature \( T_2 \) = -62°C = -62 + 273.15 = 211.15 K
Now, rearrange the Combined Gas Law to solve for \( P_2 \):
[tex]\[
P_2 = \frac{P_1 V_1 T_2}{T_1 V_2}
\][/tex]
Substitute the known values into the equation:
[tex]\[
P_2 = \frac{805 \text{ mmHg} \times 2.19 \text{ L} \times 211.15 \text{ K}}{320.15 \text{ K} \times 9.88 \text{ L}}
\][/tex]
Now calculate the numerator and the denominator separately:
Numerator:
[tex]\[
805 \times 2.19 \times 211.15 \approx 371849.8675
\][/tex]
Denominator:
[tex]\[
320.15 \times 9.88 \approx 3161.882
\][/tex]
Now divide the numerator by the denominator to find \( P_2 \):
[tex]\[
P_2 \approx \frac{371849.8675}{3161.882} \approx 117.68 \text{ mmHg}
\][/tex]
Thus, the pressure inside the balloon when it is cooled to -62°C and expands to 9.88 L is [tex]\( \boxed{117.68 \text{ mmHg}} \)[/tex].
