A balloon used for atmospheric research has a volume of 1.0 x 106 L. Assume that

Question

the balloon is filled with helium gas at STP (standard temperature and pressure) and
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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?)

Answer 1

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.