Question

Sixty liters of a gas were collected over water when the barometer read 663 mmhg , and the temperature was 20∘c. what volume would the dry gas occupy at standard conditions? (hint: consider dalton's law of partial pressures.)

Answer 1

First, let's compute the number of moles in the system assuming ideal gas behavior. 

PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles

At standard conditions, the standard molar volume is 22.4 L/mol. Thus,

Standard volume = 22.4 L/mol * 2.176 mol = 48.74 L

Answer 2

At standard conditions the volume occupied by gas is 48.74L.

At STP molar volume of gas is 22.4 L

Assuming ideal condition STP

PV= nRT

$\rm \bold{ (663mmHg)(760mmHg) (60L)= n (0.08210) (293K)}\\\rm \bold{ n = 2.176 mole}$

Hence,

$\rm \bold {Standard Volume =22.4L/mol\times 2.176mol}\\\\\rm \bold {Standard Volume = 48.74 L}$

Hence we can conclude that at standard conditions the volume occupied by gas is 48.74L.

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