Can anyone solve this?

Question

How many oxygen molecules are in 22.4 liters of oxygen gas at 273 K and 101.3 kPa?
A)6.023 × 1023
B)1.2046 × 1024
C)2.24 × 10234
D)1.3491 × 1025

Answer 1

Christoph,

 

This is a question about the ideal gas law.  We are given pressure, volume of the container, and temperature and we are asked to solve for molecules.  First, we'll solve for n (moles) and then convert that to molecules.

 

The gas law equation is PV = nRT which can be re-arranged to

 

n= PV/RT

 

where P (pressure) is in units of atm

          V (volume) is in units of liters

          T (temperature) is in Kelvin

          R (the universal gas constant) is 0.08206 L-atm/ mole-K

 

The pressure is given as 101.3 kPa.  1.0 atm = 101.3 kPa so that's what we'll plug in for P

 

So,

 

n= (1 atm)(22.4L)/(0.08206 L-atm/mol-K)(273 K)

 

n=0.99 mole, but we have to convert this to molecules. Since one mole is equal to 6.02 x10∧23, the answer is A.

 

 

Answer 2

P V = n R T    [Piv-nert]

 

(Pressure) * (Volume) = (Number of moles) * (Gas Constant, R) * (Temperature)

http://en.wikipedia.org/wiki/Gas_constant

 

 

(101.3 kPa) * (22.4 L) = n * (8.3144622 J/(mol K)) * (273 K)

 

Solve for n, make sure the units are converted where they need to be.

Answer 3

Using the Ideal Gas Law, PV = nRT:

 

n = PV/RT = (101.3 kPA*22.4L)/(8.31 J/(mol K) * 273 K) =

= (101.3 * 1000 N/m^2 * 22.4/1000 m^3)/(8.31 N m/(mol K) * 273 K) =

= (101.3 * 22.4 N m)/(8.31 * 273 N m/mol) =

= (101.3 * 22.4/(8.31*273)) mol =

= 1.00022 mol

 

Number of molecules = n*6.02*10^23 molecules/mol = 6.021 * 10^23

 

This is very close to answer A. It is slightly different from the answer, because of the different precisions of the values of the constant used.