How can I calculate final pressures in a reversible argon expansion?

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marissy
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isothermal and adiabatic please help!

Homework Statement


Hello,

I have been stuck on this problem for about an hour so any hep would be greatly appreciated!

One mole of argon at 25 degrees C and 1 atm pressure is allowed to expand reversibly t a volume of 50 L (a) isothermally (b) adiabatically. Calculate the final pressure in each case assuming ideal gas behavior.
The answers have been given and they are: (a) 0.489 atm and (b) 0.303 atm

I can't get either a or b for some reason I don't know what I am doing wrong! Please help me :(


Homework Equations


Cv = 3/2 R Cp= 5/2 R
Boyle's law


The Attempt at a Solution



my answers don't even come close...?
 
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Show some work please. It's impossible to show you where you went wrong if you don't show us what you have done.
 


nevermind i got it all -- I just didnt have my P and V in the right units... anyway i found V1 using the ideal gas law, and the used boyle's law to find P2. Seems so simple when you have the correct units :)

for part b i did PV(5/3)=nRT to find V1 and then used boyle's law to find P2. I got 0.296 for the answer tough. maybe my sig. figs. are off? i can't find out why.
 


marissy said:
for part b i did PV(5/3)=nRT to find V1 and then used boyle's law to find P2. I got 0.296 for the answer tough. maybe my sig. figs. are off? i can't find out why.
You will have to use the adiabatic condition:

[tex]PV^\gamma = \text{constant} \ne nRT[/tex]

for part b). [itex]PV^1 = nRT[/itex]

AM
 


Yeah, as AM has said, use the adiabatic condition, along with energy conservation to figure it out.