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March 3, 2015

March 3, 2015

Posted by **Anonymous** on Sunday, January 9, 2011 at 10:58am.

a) at STP

b) at 100 atm pressure and 0 degrees Celsius

((Assume for simplicity that the ideal-gas equation still holds.))

Ar

a= 1.34

b = 0.0322

I'm extremely confused about this question, are they asking for the volume when they state fraction of the volume.

- Chemistry -
**drwls**, Sunday, January 9, 2011 at 11:52amThe units of b are liters/mole. So the actual molecules in a liter of Argon occupy b/4 or 0.008 liters.

At STP, the (mostly empty) space occupied by gas is 22.4 liters.

For the fraction, take the ratio.

Then do the same thing for the 100 atm case. The volume occupied (including space between atoms) will be 100 times less than at 1 atm STP, or 0.224 liters.

- Chemistry -
**Anonymous**, Sunday, January 9, 2011 at 11:59amTherefore, they are asking for the volume

(a) at STP

v = 22.4 L

T = 273 K

P = 1 atm

Would I use Van der Waals equation for (a) or for (b)? I think I would have to use this equation for b

(a) V = nRT/ P

Can I use this for question a

- Chemistry -
**Anonymous**, Sunday, January 9, 2011 at 12:10pmOkay, then what equation can I use for (a)

Instead of using the ideal-gas equation,

can I use the mole-fraction

By stating "For the fraction, take the ratio. " does this apply to the mole-fraction

- Chemistry -
**drwls**, Sunday, January 9, 2011 at 12:25pmYou should take the ratio of the computed volumes.

0.008/22.4 in the first case

It doesn't make sense to talk about the mole fraction of solid spheres

- Chemistry -
**Anonymous**, Sunday, January 9, 2011 at 12:29pmOkay but wait we can find the number of moles using PV = nRT? Right?

V = 22.4 L

P = 1 atm

R= 0.0831 L-atm

T = 273 K

- Chemistry -
**kk**, Tuesday, October 23, 2012 at 6:59amhaha

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