Air is a mixture of many gases. a 25.0-liter jar of air contains 0.0104 mole of argon at a temperature of 273 k. what is the partial pressure of argon in the jar?

 The  partial  pressure  of argon  in the jar is 0.0093 atm

  calculation

This  is calculated using the ideal gas equation

that is PV =nRT

where;

P(pressure) =?

V ( volume)= 25.0 L

n(number of moles) =0.0104 moles

R( gas constant) = -0.0821  L. atm/ mol.K

T( temperature) = 273 K

make p the subject of the formula by diving both side of equation by v

P = nRT/V

P =[(0.0104 mol x 0.0821 L.atm/mol.k  x 273 k)/ 25 L ] = 0.0093 atm

0.944

Explanation:

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