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

2 Answers
Jan 14, 2018

Answer:

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#....

Jan 14, 2018

Answer:

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#