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

11.427L

Explanation:

From the ideal gas law,

PV=nRT

Where;

P=pressure of the gas

V=volume of the gas occupied by the container

n=number of moles of the gas

R=the ideal gas constant=0.0821atmL/mol/K

Given

P=4.258atm

V=?

n=2.0801mol

T=284.92K

From the equation

PV=nRT,

We will make V subject of formula

V=nRT/P

Let's now substitute the values

V = (2.0801*0.0821*284.92) / 4.258

V=11.427L

Therefore, the volume of the container is 11.427L