Calculate the number of atoms of argon in 75.0 L of argon at STP.

Answer 1

Because we are told to consider the Argon gas at STP (standard temperature and pressure),

we are able to assume that Ar behaves like an ideal gas.

Therefore, we know

1. volume of 1 mole of gas at STP conditions is 22.4 L (for an ideal gas)

DETERMINE HOW MANY MOLES OF GAS IN 75 L

75 L x 1 mol ~= 3.348 mol Ar (g)

22.4 L

CONVERT MOLES TO ATOMS USING AVAGADRO'S NUMBER

3.348 mol Ar x 6.02x10²³ atoms Ar =

1 mol Ar

ANSWER:

= 2.0x10²⁴ atoms Ar

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CONFIRM USING THE IDEAL GAS LAW AT STP:

pV = nRT, where R = 0.08206 atm⋅L/(mol⋅K)

1. standard Temperature is 0 °C = 273 K (always use Kelvin for gas)
2. standard Pressure is 1 atm

(same amount in moles will yield the same number of atoms)

n = pV → 1atm⋅75L ~= 3.348 mol Ar (g) √

RT 0.08206⋅273K