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?

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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?

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Answer 1 · 2015-11-23T02:03:16

The volume of argon will be 6.96 L.

Explanation:

Use the ideal gas law $P V = n R T$ $PV=nRT$ , where $n$ $n$ is moles and $R$ $R$ 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.

$Ar :$ $"Ar":$ $10.8 g Ar \times \frac{1 mol Ar}{39.948 g Ar} = 0.27035 mol Ar$ $10.8"g Ar"xx(1"mol Ar")/(39.948"g Ar")="0.27035 mol Ar"$

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

Unknown
$V$ $V$

Equation
$P V = n R T$ $PV=nRT$

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

$V = \frac{n R T}{P}$ $V=(nRT)/P$

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

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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?

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