A weather balloon is filled with helium that occupies a volume of 5.75 104 L at 0.995 atm and 32.0°C. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -14.7°C. What is the volume of the balloon at that new location?
the volume is directly proportional to the absolute (Kelvin) temperature
... and inversely proportional to the ambient pressure
To find the volume of the balloon at the new location, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (0.995 atm)
V1 = initial volume (5.75 x 10^4 L)
T1 = initial temperature (32.0°C + 273.15 = 305.15 K)
P2 = final pressure (0.720 atm)
V2 = final volume (unknown)
T2 = final temperature (-14.7°C + 273.15 = 258.45 K)
Now we substitute the values into the equation and solve for V2:
(0.995 atm * 5.75 x 10^4 L) / (305.15 K) = (0.720 atm * V2) / (258.45 K)
Cross-multiplying and rearranging the equation, we get:
(0.995 atm * 5.75 x 10^4 L * 258.45 K) = (0.720 atm * V2 * 305.15 K)
Dividing both sides by (0.720 atm * 305.15 K):
(V2) = [(0.995 atm * 5.75 x 10^4 L * 258.45 K)] / (0.720 atm * 305.15 K)
Calculating the equation:
V2 = [3724574.875 atm L K] / [219888.08 atm K]
V2 ≈ 16.95 x 10^3 L
Therefore, the volume of the balloon at the new location is approximately 16.95 x 10^3 L.