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
Name(s):________________________________ CHM 1020 W7
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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 variables given are pressure, volume, and temperature. To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
Since the number of moles of gas remains constant in this scenario, we can rewrite the equation as:
P1V1/T1 = P2V2/T2
Where:
P1 = initial pressure (STP = 1 atm)
V1 = initial volume (1.0 x 10^6 L)
T1 = initial temperature (273 K or 0°C)
P2 = final pressure (243 mm Hg converted to atm)
V2 = final volume (unknown)
T2 = final temperature (-33°C converted to Kelvin)
Converting the final pressure to atm:
243 mm Hg * (1 atm / 760 mm Hg) = 0.3197 atm
Converting the final temperature to Kelvin:
-33°C + 273 = 240 K
Now we can plug in the values into the equation:
(1 atm)(1.0 x 10^6 L) / (273 K) = (0.3197 atm)(V2) / (240 K)
V2 = (1 atm)(1.0 x 10^6 L)(240 K) / (273 K)(0.3197 atm)
V2 = 830,724 L
Therefore, the volume of the balloon under the given atmospheric conditions will be approximately 830,724 L.