## Quiz 1 #3, reversible isothermal

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

Navjot 1G
Posts: 1
Joined: Tue Nov 25, 2014 3:00 am

### Quiz 1 #3, reversible isothermal

I am having trouble with problem #3:
"If 2.00 mol of an ideal gas at 300K and 3.00 atm expands isothermally and reversibly from 6.00L to 18.00 L has a final pressure of 1.20 atm, what is w, q, and U?"

I am not sure which formula to use to calculate work. I used the (vf/vi) and (pi/pf) formula but I got different numbers and both are very large (-5480.31 and -4570.82). Which am I supposed to use? Are these values relatively correct? Is there a type of max/min number that work can be?

Also, would deltaU always equal zero for a reversible isothermal expansion?

Thank you!

Kendal Reeder 1E
Posts: 15
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Quiz 1 #3, reversible isothermal

Yes, delta U always equals zero for an isothermal reversible expansion. Because delta U = q + w, q + w = 0 and q = -w. For a reversible expansion, use the formula w = -nRTln(Vf/Vi). Hope this helps!

martha-1I
Posts: 76
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Quiz 1 #3, reversible isothermal

You do not need to calculate the work with respect to pressure because the equation w=-nRTln(v2/v1) is derived from w=-P$\Delta$V. In the derivation we substituted nRT/v for P because P is related to PV=nRT. We then integrated and ended up with the equation w=-nRTln(v2/v1). This derivation lets us know that nRT in the equation of work accounts for change in pressure as well as for volume so finding work with respect to pressure would be redundant and unnecessary.