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September 2, 2014

September 2, 2014

Posted by **tom** on Thursday, October 16, 2008 at 1:42am.

(a) Determine the initial volume in m^3.

i tried using the formula w=integral f/i pdv

3400J= -(1.00atm)(.022m^3Vf-Vi)

i eneded with 1.55 x10^5 but htis is not the right answer. HELP PLS!

- physics -
**Damon**, Thursday, October 16, 2008 at 5:16amIsothermal expansion

W = integral p dv

p = n R T/v, T constant so

W = n R T integral dv/v

W = n R T ln v from v1 to v2

W = n R T ln(v2/v1)

here

n = 1

R = R =.08206 L atm/mol degK or 8.31 J/mol deg K

Need T, use v2

T = p2 v2/nR = 1*22/1*.082

T = 268 deg K

now we want to use Joules and scientific notaton R

V2 = 22 L = 22* 10^-3 m^3

3400 = 1 * 8.31 * 268 ln (22*10^-3/v)

ln(.022/v)= 1.53

.022/v = 4.6

v = .00478 m^3 = 4.78 Liters

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