Question

Calculate the volume occupied by 2.00 mol of oxygen gas at 273 k and 1.00 atm

Answer 1

Step 1

Oxygen gas is assumed to be an ideal gas, so it is applied:

`p\text{ x V = n x R x T}`

p = pressure

V = volume

n = number of moles

R = gas constant

T = absolute temperature

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Step 2

Information provided:

V = unknown

p = 1.00 atm

n = 2.00 moles

T = 273 K

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Information needed:

R = 0.082 atm L/mol K

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Step 3

From step 1, V is found as follows:

`\begin{gathered} p\text{ x V = n x R x T} \\ V\text{ = }\frac{n\text{ x R x T}}{p} \\ V\text{ = }\frac{2.00\text{ moles x 0.082 }\frac{atm\text{ L}}{mol\text{ K}}\text{ x 273 K}}{1.00\text{ atm }} \\ V\text{ = 44.77 L = 44.8 L \lparen apporox.\rparen} \end{gathered}`

Answer: 44.8 L