Question

Calculate q, w, ∆U and ∆H for 1.00 mole of a monatomic ideal gas which undergoes a change of state along two different paths. (Note that the initial and final states are the same for both paths.) Report all energies in joules.

Path I: Reversible isothermal compression at T= 298 K from 1.00 bar to 2.00 bar.

Path II: (2 steps)

1.       Isobaric (constant pressure) cooling of the gas at 1.00 bar until the volume of the gas is equal to the final volume in Path I.

2.       Constant volume heating of the gas at this new volume to its original temperature.

Sketch the 2 paths on a P-V diagram

Answer 1

for isothermal expansion of an ideal gas, change in internal energy=0 and deltaH=0 and

from 1st law, deltaU= Q+W, W= -Q= -nRT*ln(P1/P2)=-1*8.314*298*ln(1/2)=1717 joules

Q=-1717 joules

from gas law, initial volume= nRT/P= 1*0.0821L.atm/mole.K*298/1= 24.5 L

from P1V1= P2V2, V2 at constant temperature, V2= final volume= P1V1/P2= 1*24.5/2 = 12.25 L

2. for second path at constant pressure, V1/T1= V2/T2

24.5/298= 12.25/T2

T2= 12.25*298/24.5= 149K

work done, W= -nR*deltaT= -8.314*(149-298)=1239 joules, deltaU=nCv*dT= 1*1.5*8.314*(149-298)= -1858 joules

Q= deltaH= nCp*deltaT= 1*2.5*8.314*(149-298)= -3097 joules

2. for constant volume heating work done =0, deltaU= Q= nCV*deltaT= 1*1.5*8.314*149=1858 joules

deltaH= nCp*deltaT= 2.5*8.314*149=3097 joules

for overall , deltaU= 1858-1858=0, deltaH= 3097-3097=0, W= 1239 joules, Q=-3097+1858=-1239 joules

the PV diagram is shown below