Thermodynamics – reversible isothermal cycle

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I can do everything else except work for part one going from A to B.

What I did was

-∫PdV = -nRT ∫(1/v)dV = -nRT ln (Vf/Vi)

I can solve for T because everything is given, T in Kelvin is 882.
From PV=nRT
T=(5atm)(10L)/1mol*R = 882K

where R = .0820582

Therefore w= -1mol * R * 882 * ln (5) = – 11.8kJ

where R = 8.314

But this is incorrect. It is supposed to be -8.15kJ

http://ift.tt/Q6FGwE

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