Question #8dabc

3 Answers

The answer is : V = 2.80x10^-4L.

Gas' properties:
Volume (V)
Pressure (P)
Temperature (T- in Kelvins)
Amount (n - moles)
.
And a Constant R (0.0821 L x atm / mol x K)

The easiest way to solve this problem is to make a set up as follow: !!! STP !!!
V= ? L
P= 1 atm
T= 273 K
n= ?

1) We can get the amount of moles from 5.5104g of CO2 gas.

MolarmassCO2=12.01+216=44.01g of CO2

Now take :

5.5104gCO21moleCO244.01gCO2=1.25105 mol of CO2

2) PV=nRT

1V=1.251050.0821273
V=2.80104L

And that's it !

I strongly hope I was helpful !

David Tran
trananhdavid@gmail.com

Dec 16, 2014

You would use the ideal gas law in order to solve this problem. The equation for the ideal gas law is PV = nRT. STP for the gas laws is 0oC and 1 atm. The temperature must be converted to Kelvins, and the mass of CO2 must be converted to moles.

Given/Known:
P = 1 atm
molar mass of CO2 = 44.01 g/mol
n = 5.5 x 104g x 1 mol CO244.01 g CO2 = 1.2 x 105mol CO2
R = 0.08205746 L atm K1mol1
T = 0oC+273.15=273.15K

Unknown:
V

Equation:
PV = nRT

Solution: Divide both sides of the equation by P, to isolate V. Solve for V.

V = nRTP = 1.2 x 105x 0.08205746 x 273.15/1 = 2.7 x 104L (Units removed in order to make the equation more compact.)

Answer:
The volume of 5.5 x 104g CO2 at STP is 2.7 x 104L.

Dec 16, 2014

An alternative approach to this problem is by using molar volume at STP.
We know that, at STP, one mole of any ideal gas occupies 22.4L.

The number of moles of CO2, knowing that its molar mass is 44.01gmol, is

nCO2=mCO2molarmass=5.5104g44.01gmol=1.2105 moles

Therefore, since n=VVmolar, we get

V=nCO2Vmolar=1.2105moles22.4L1mole=2.7104L

One can use this method as a primary tool or as a way to double-check the result determined using the ideal gas law.