A weather balloon rises through the atmosphere, its volume expanding from
to20 to 10
as the temperature drops from
to 1.0
If the initial gas pressure inside the balloon is 4.0 to12m
what is the final pressure
The Correct Answer and Explanation is:
To solve this problem, we apply the Combined Gas Law, which relates the pressure, volume, and temperature of a gas:P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}T1P1V1=T2P2V2
Where:
- P1P_1P1, V1V_1V1, and T1T_1T1 are the initial pressure, volume, and temperature,
- P2P_2P2, V2V_2V2, and T2T_2T2 are the final pressure, volume, and temperature.
Step 1: Write down known values
Let’s assume your incomplete numbers were meant to be:
- Initial volume: V1=2.0 LV_1 = 2.0 \, \text{L}V1=2.0L
- Final volume: V2=10.0 LV_2 = 10.0 \, \text{L}V2=10.0L
- Initial temperature: T1=20∘C=293 KT_1 = 20^\circ\text{C} = 293 \, \text{K}T1=20∘C=293K
- Final temperature: T2=1.0∘C=274 KT_2 = 1.0^\circ\text{C} = 274 \, \text{K}T2=1.0∘C=274K
- Initial pressure: P1=4.0×102 mmHgP_1 = 4.0 \times 10^{2} \, \text{mmHg}P1=4.0×102mmHg
- Final pressure: P2=?P_2 = ?P2=?
Step 2: Rearrange the Combined Gas Law to solve for P2P_2P2:
P2=P1V1T2V2T1P_2 = \frac{P_1 V_1 T_2}{V_2 T_1}P2=V2T1P1V1T2
Step 3: Plug in the values:
P2=(4.0×102)(2.0)(274)(10.0)(293)P_2 = \frac{(4.0 \times 10^2)(2.0)(274)}{(10.0)(293)}P2=(10.0)(293)(4.0×102)(2.0)(274)P2=(800)(274)2930=2192002930≈74.8 mmHgP_2 = \frac{(800)(274)}{2930} = \frac{219200}{2930} \approx 74.8 \, \text{mmHg}P2=2930(800)(274)=2930219200≈74.8mmHg
✅ Final Answer:
The final pressure is approximately 74.8 mmHg\boxed{74.8 \, \text{mmHg}}74.8mmHg.
📘 Explanation (Like in textbooks):
When a weather balloon rises through the atmosphere, it encounters lower external pressure and lower temperatures. According to the Combined Gas Law, the relationship between pressure (P), volume (V), and temperature (T) for a fixed amount of gas is:P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}T1P1V1=T2P2V2
This equation shows that if the temperature decreases and the volume increases, the pressure must decrease. Physically, this makes sense: as the balloon rises and expands due to lower atmospheric pressure, the internal gas cools and exerts less pressure on the balloon walls.
This principle is fundamental in understanding gas behavior in varying atmospheric conditions and is critical in meteorology and high-altitude physics.
