Question

What volume will `5.2 xx 10^24` molecules of `CO_2` occupy?

Answer 1

Well, clearly a pellet of dry ice will occupy LESS volume than carbon dioxide gas....

Explanation:

And so we specify conditions of `P=1*atm`, and `T=298*K`...and to answer your question MEANINGFULLY, we specify that `5.2xx10^24*"molecules of dry ice"`, were confined in a piston, and the gas expanded against `1*atm....`

And so............ `V=(nRT)/P=((5.2xx10^24*CO_2)/(6.022xx10^23*mol^-1)xx0.0821*(L*atm)/(K*mol)xx298*K)/(1*atm)`..

And I make this over `200*L`....

Answer 2

At what pressure and temperature?
Do you mean at STP?

Explanation:

We can use the ideal gas equation: PV = nRT
(Pressure)(Volume) = (moles)(gas constant)(Temperature)

You have given me these parameters:
moles = `(5.2*10^24) * (1mol)/(6.022*10^23) = 8.635molCO_2`
gas constant = `(0.0821(L)(atm))/((mol)(K))`
volume = x(variable)
pressure = ?(not given)
temperature = ?(not given)

If you mean at STP, where the pressure is 1 atm and the temperature is 273K, here is the equation and answer:
`(1atm)(xL) = (8.635molCO_2)((0.0821*L*atm)/(mol*K))(273K)`
`x ~~ 193.5388`
Significant Digits:
`color(blue)(x ~~ 190L`