how many atoms of argon gas are in 137 ml container at the pressure in the container is 8.80 X 10^5 mmhg and the temperature to 794 K​

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#Ar atoms = 1.47 x 10²²

Explanation:

PV = nRT => n = PV/RT

P = 8.80 X 10⁵ mmHg =  8.80 X 10⁵ mmHg/760mmHg·atm⁻¹ = 11.58atm

V = 137ml = 0.137 Liter

n = ? moles Ar = ? moles Ar x 6.02 x 10²³ Atoms Ar/mole Ar

R = 0.08206 L·atm/mol·K

T = 794K

PV = nRT => n = PV/RT = 11.58atm·0.137L / 0.08206L·atm/mol·K·794K

=> n = 0.0243mole Ar = 0.0244mole Ar x 6.02 x 10²³ atoms Ar/mole Ar

= 1.47 x 10²² atoms Ar

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