A Partially weather balloon has a volume of 195 L at a pressure of 0.993 atm. What will be the pressure in the balloon when it is inflated to a volume of 7500 L? (assume to temp. change)
p1v1 = p2v2
Use Boyles Law if mass of gas and temperature are unchanged. (P1)(V1) = (P2)(V2). However, you must give temperature change if that is to be taken into account. On the average, temperature of the atmosphere drops 2-deg Celcius per 1000-ft of elevation. Just using the data given and Boyles Law, the atmospheric pressure would have decreased to 0.026 atm in order for balloon to increase volume to 7500L. Base on standard pressure lapse rate at ~0.001 atm/1000 ft => Altitude of balloon ~ 30,000 ft. A std temp laps rate of 2-deg Celcius / 1000 ft => Temp ~ -60-deg Celcius from surface temp. Assuming surface temp is ~25-deg Celcius to a Temp of -60-deg celcius, the pressure would drop to ~0.019-atm at 30,000ft where temps are ~-60-deg Celcius.
To solve this problem, we can use the combined gas law, which states that the product of the initial pressure and initial volume divided by the initial temperature is equal to the product of the final pressure and final volume divided by the final temperature.
The formula for the combined gas law is:
(P1 * V1) / T1 = (P2 * V2) / T2
In this case, we don't have information about the temperature, but the problem states that we can assume no temperature change. Therefore, the temperature can be canceled out in our equation.
Let's label our given values:
P1 = 0.993 atm (initial pressure)
V1 = 195 L (initial volume)
V2 = 7500 L (final volume)
P2 = ? (final pressure)
Now we can substitute these values into the formula:
(0.993 atm * 195 L) / T1 = (P2 * 7500 L) / T2
Since we are assuming no temperature change, T1 = T2, so the temperature cancels out:
0.993 atm * 195 L = P2 * 7500 L
To solve for P2, we need to isolate it on one side of the equation. Divide both sides of the equation by 7500 L:
(0.993 atm * 195 L) / 7500 L = P2
Now, the units cancel out, and we can calculate P2:
P2 = (0.993 atm * 195 L) / 7500 L
P2 ≈ 0.0258 atm
Therefore, the pressure inside the balloon when inflated to a volume of 7500 L will be approximately 0.0258 atm.