Using The Ideal Gas Law Equation (PV=nRT), What Would The Theoretical Volume Of Oxygen Can That Can Be Generated? Assuming P=1.00 Atm, T=298 K, And R=0.0821L-atm/mol-K


2 Answers

Ken Moorhead Profile
Ken Moorhead answered
Is a value for n (number of mols of gas) given?

If you know a value for n, V=nRT/P

If no value for n is given, I would assume you have one mol; but you should also check with the instructor. There are other less-common forms of the ideal gas law, but I don't believe they are used much in chemistry courses.
eye3 light Profile
eye3 light answered
Oxygen is in the gaseous form so considering it ideal gas ,we can use ideal gas law. but here it is not mentioned in the statement that what is the mass or moles of the oxygen. so first of all we need the mass of the oxygen.


First of all assume that the moles of oxygen = n =1
Given in the statement that pressure = P= 1 atm
Temperature =T =298 K
Gas constant = R= 0.082L-atm/mol-K

Ideal Gas law
V= (1mole)( 0.082L-atm/mol-K)(298 K)/(1 atm)
V =24.236 Liter
This above volume is for 1 mole of oxygen
So Volume of oxygen is 24.236 Liter per Mole
But if you know the quantity of oxygen in Moles then you can find out the Volume for that quantity of oxygen at the above mention conditions by the formula

Formula 1 (If quantity is in moles or kilomoles)

V = (24.236 n) liter
n = number of moles of oxygen
Formula 2 (If quantity is in grams of kilo grams)
V=(24.236 m/M)
V=(24.236 m/32)
m=mass of the oxygen
M= Molecular weight of oxygen =32

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