- #1

Rizke

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Does not use the template because it was originally posted in a non-homework forum

A welder using a tank of volume 7.50×10^−2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.00×10^5 Pa and temperature of 36.5 ∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.0 ∘C, the gauge pressure of the oxygen in the tank is 1.85×10^5 Pa . Find the initial mass of oxygen.

I can't seem to get this right. i even worked backwards with that answer and still. I read in a similar thread that i need the absolute value of gas pressure. but not sure how to apply

My Attempt: PV=nRT

P = 3.0e^5 - in pascals

V = 7.5e^-2

R = 8.3144621

T = (36.5 C + 273.15)= 309.65 K

(3.0e^5)(7.5e^-2) / (8.3144621)(309.65) = n => n= 8.7393 moles

mass = 8.7393* 32 = > 279.658 grams. but .280 kg is not .374 kg.

So yeah I am puzzled.

*correct answer : .374 kg*I can't seem to get this right. i even worked backwards with that answer and still. I read in a similar thread that i need the absolute value of gas pressure. but not sure how to apply

My Attempt: PV=nRT

P = 3.0e^5 - in pascals

V = 7.5e^-2

R = 8.3144621

T = (36.5 C + 273.15)= 309.65 K

(3.0e^5)(7.5e^-2) / (8.3144621)(309.65) = n => n= 8.7393 moles

mass = 8.7393* 32 = > 279.658 grams. but .280 kg is not .374 kg.

So yeah I am puzzled.

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