Question

What is the volume occupied by 10.8 g of argon gas at a pressure of 1.30 atm and a temperature of 408 K?

Answer 1

The volume of argon will be 6.96 L.

Explanation:

Use the ideal gas law `PV=nRT`, where `n` is moles and `R` is the gas constant.

Since we have mass and need moles, we must divide the given mass of argon by its molar mass (atomic weight on the periodic table in g/mol). The molar mass of argon is 39.948 g/mol.

`"Ar":` `10.8"g Ar"xx(1"mol Ar")/(39.948"g Ar")="0.27035 mol Ar"`

I am keeping a couple of guard digits to reduce rounding errors. I will round the final answer to three significant figures.

Ideal Gas Law

Given/Known
`P="1.30 atm"`
`n="0.27035 mol"`
`R="0.082057 L atm K"^(-1) "mol"^(-1)"`
https://en.wikipedia.org/wiki/Gas_constant
`T="408 K"`

Unknown
`V`

Equation
`PV=nRT`

Solution
Rearrange the equation to isolate volume, `V`, and solve.

`V=(nRT)/P`

`V=(0.27035cancel"mol"xx0.082057"L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1))xx408cancel"K")/(1.30cancel"atm")="6.96 L Ar"`