Answer :
To determine the air pressure at the top of the mountain, we can use the ideal gas law relation in the form:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Here's the step-by-step process to solve for [tex]\( P_2 \)[/tex]:
1. Identify the known values:
- [tex]\( P_1 = 1.0 \text{ atm} \)[/tex] (Pressure at the valley floor)
- [tex]\( V_1 = 0.985 \text{ L} \)[/tex] (Volume of the bag at the valley floor)
- [tex]\( T_1 = 25^{\circ} C \)[/tex]
- [tex]\( V_2 = 1.030 \text{ L} \)[/tex] (Volume of the bag at the top of the mountain)
- [tex]\( T_2 = 22^{\circ} C \)[/tex]
2. Convert temperatures from Celsius to Kelvin:
- [tex]\( T_1 = 25 + 273.15 = 298.15 \text{ K} \)[/tex]
- [tex]\( T_2 = 22 + 273.15 = 295.15 \text{ K} \)[/tex]
3. Rearrange the formula to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2} \][/tex]
4. Plug in the values:
[tex]\[ P_2 = \frac{1.0 \text{ atm} \cdot 0.985 \text{ L} \cdot 295.15 \text{ K}}{298.15 \text{ K} \cdot 1.030 \text{ L}} \][/tex]
5. Calculate [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{0.985 \cdot 295.15}{298.15 \cdot 1.030} \][/tex]
[tex]\[ P_2 \approx 0.9466882344034164 \text{ atm} \][/tex]
6. Apply significant figures:
Since the given values [tex]\( V_1 \)[/tex], [tex]\( V_2 \)[/tex], and [tex]\( P_1 \)[/tex] all have 3 significant figures, the answer should also be reported with 3 significant figures.
Thus, the air pressure at the top of the mountain is:
[tex]\[ \boxed{0.947 \text{ atm}} \][/tex]
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Here's the step-by-step process to solve for [tex]\( P_2 \)[/tex]:
1. Identify the known values:
- [tex]\( P_1 = 1.0 \text{ atm} \)[/tex] (Pressure at the valley floor)
- [tex]\( V_1 = 0.985 \text{ L} \)[/tex] (Volume of the bag at the valley floor)
- [tex]\( T_1 = 25^{\circ} C \)[/tex]
- [tex]\( V_2 = 1.030 \text{ L} \)[/tex] (Volume of the bag at the top of the mountain)
- [tex]\( T_2 = 22^{\circ} C \)[/tex]
2. Convert temperatures from Celsius to Kelvin:
- [tex]\( T_1 = 25 + 273.15 = 298.15 \text{ K} \)[/tex]
- [tex]\( T_2 = 22 + 273.15 = 295.15 \text{ K} \)[/tex]
3. Rearrange the formula to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2} \][/tex]
4. Plug in the values:
[tex]\[ P_2 = \frac{1.0 \text{ atm} \cdot 0.985 \text{ L} \cdot 295.15 \text{ K}}{298.15 \text{ K} \cdot 1.030 \text{ L}} \][/tex]
5. Calculate [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = \frac{0.985 \cdot 295.15}{298.15 \cdot 1.030} \][/tex]
[tex]\[ P_2 \approx 0.9466882344034164 \text{ atm} \][/tex]
6. Apply significant figures:
Since the given values [tex]\( V_1 \)[/tex], [tex]\( V_2 \)[/tex], and [tex]\( P_1 \)[/tex] all have 3 significant figures, the answer should also be reported with 3 significant figures.
Thus, the air pressure at the top of the mountain is:
[tex]\[ \boxed{0.947 \text{ atm}} \][/tex]
