Prove non adiabatic reversible expansion and ΔS

  • Thread starter Nick_273
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  • #1
Nick_273
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Homework Statement


One mole of a monatomic ideal gas is transformed from 0.00°C and 2.00 atm to
-40.0°C and 0.400 atm. Show that this is NOT an adiabatic reversible expansion and
calculate ΔS for the change. (Hint: you will need to consider a reversible path that
includes a constant pressure step and a constant temperature step.)


Homework Equations


ΔU=q+w
pV=nRT
ΔS=nC(p,m)ln(Vf/Vi)

The Attempt at a Solution


For adiabatic reversible expansion, q=0, so i have to show that q= a number.
Found Vi= 11.2L and Vf=47.8L via ideal gas law

ΔS=(1mol)(5/2(8.314J/Kmol)ln(47.2/11.2)
=30.15J/K
 
Last edited:

Answers and Replies

  • #2
I like Serena
Homework Helper
MHB
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250
Hi Nick_273! :smile:

I'm afraid it does not suffice to calculate ΔS for the change in volume.
You also need to account for the change in either pressure or temperature.

There's a few more relevant equations that you could derive, but you can also find them here:
http://en.wikipedia.org/wiki/Table_of_thermodynamic_equations

In particular you can find the equations for ΔS here when p is constant, but also when T is constant.
You need these 2 to follow the hint.
 

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