. A balloon used for atmospheric research has a volume of 1.0 x 106 L. Assume that the balloon is filled with helium gas at STP (standard temperature and pressure) and then allowed to ascend to an altitude of 10 km, where the pressure of the atmosphere is 243 mm Hg and the temperature is -33°C. What will the volume of the balloon be under these atmospheric conditions? (Hint: Three variables have been given so what equation will you use?)
The ideal gas law can be used in this scenario. The ideal gas law is represented by the equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
Since the number of moles of gas remains constant, we can rewrite the ideal gas law as:
P1V1/T1 = P2V2/T2
Where:
P1 = initial pressure at STP (1 atm)
V1 = initial volume at STP (1.0 x 10^6 L)
T1 = initial temperature at STP (273 K)
P2 = final pressure at 10 km altitude (243 mm Hg)
V2 = final volume (what we are trying to find)
T2 = final temperature at 10 km altitude (-33°C = 240 K)
Now we can plug in the values and solve for V2:
(1 atm)(1.0 x 10^6 L)/(273 K) = (243 mm Hg)(V2)/(240 K)
V2 = (1 atm)(1.0 x 10^6 L)(240 K)/(273 K)(243 mm Hg)
V2 = 847,953 L
Therefore, the volume of the balloon at 10 km altitude will be approximately 847,953 L.