2.0801 mol of bromine gas is held at 4.258 atm and 284.92 K. What is the volume of its container in liters?

11.41L

Explanation:

Data obtained from the question include:

n (number of mole of Br2) = 2.0801 mol

P (pressure) = 4.258 atm

T (temperature) = 284.92K

R (gas constant) = 0.082atm.L/Kmol

V (volume of the container) =?

Applying the ideal gas equation PV = nRT, the volume of the container can be obtained as follow:

PV = nRT

4.258 x V = 2.0801x0.082x284.92

Divide both side by 4.258

V = (2.0801x0.082x284.92)/4.258

V = 11.41L

Therefore, the volume of the container is 11.41L

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