Question

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

Answer 1

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

Answer 2

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.