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A 5.0 gallon container contains 15.0 moles of propane gas at room temperature (20. °C). What is the pressure of the gas? (1.0 gal = 3.79 L)


1.3 atm

19 atm

72 atm

1930 atm

Verified Answer :

19 atm
PV=nRT
P(18.95)=(15)(0.08206)(293)

Answer: The pressure of the gas is 19 atm.

Explanation: By using ideal gas equation, which is:

\(PV=nRT\)

Given conditions are:

T = 20°C = (273 + 20)K = 293K

V = 5 gallon = ( 5 × 3.97)L = 18.95L

n = 15 moles

\(R=0.082057\text{ L atm }mol^{-1}K^{-1}\)  (Gas Constant)

P = ? atm

Putting all the values in above equation, we get

\(P\times 18.95L=(15mol)(0.082057\text{ L atm }mol^{-1}K^{-1})(293K)\)

P = 19 atm

Therefore, the pressure of the gas is found to be 19 atm.

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