Certainly! Let's solve the problem step-by-step using the Combined Gas Law which relates pressure (P), volume (V), and temperature (T) of a gas sample with the equation:
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/tex]
Here are the given values:
- Initial volume [tex]\( V_1 = 54.0 \)[/tex] L
- Initial pressure [tex]\( P_1 = 759 \)[/tex] mmHg
- Initial temperature [tex]\( T_1 = 21.1^{\circ} \text{C} \)[/tex]
First, we need to convert the initial temperature from Celsius to Kelvin:
[tex]\[
T_1 = 21.1 + 273.15 = 294.25 \text{ K}
\][/tex]
Next, we are given the conditions at the higher altitude:
- Temperature [tex]\( T_2 = -3.86^{\circ} \text{C} \)[/tex]
- Pressure [tex]\( P_2 = 0.0757 \text{ atm} \)[/tex]
Again, convert the temperature [tex]\( T_2 \)[/tex] to Kelvin:
[tex]\[
T_2 = -3.86 + 273.15 = 269.29 \text{ K}
\][/tex]
Also, convert the pressure [tex]\( P_2 \)[/tex] into mmHg (since 1 atm = 760 mmHg):
[tex]\[
P_2 = 0.0757 \times 760 = 57.532 \text{ mmHg}
\][/tex]
Now, we are ready to solve for the new volume [tex]\( V_2 \)[/tex] using the Combined Gas Law:
[tex]\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\][/tex]
Rearranging the formula to solve for [tex]\( V_2 \)[/tex]:
[tex]\[
V_2 = \frac{P_1 V_1 T_2}{T_1 P_2}
\][/tex]
Substituting the given values into the equation:
[tex]\[
V_2 = \frac{759 \,\text{mmHg} \times 54.0 \,\text{L} \times 269.29 \,\text{K}}{294.25 \,\text{K} \times 57.532 \,\text{mmHg}}
\][/tex]
[tex]\[
V_2 = 651.97 \, \text{L} \quad \text{(rounded to two decimal places)}
\][/tex]
Therefore, the volume of the weather balloon at the higher altitude is:
[tex]\[
V = 651.97 \, \text{L}
\][/tex]
