Enthalpy and energy are state functions, heat and work are not; let\'s explore t

Question

Enthalpy and energy are state functions, heat and work are not; let's explore this. One mole of an ideal monatomic gas initially at 10 atm and 25 degree C is allowed to expand against a constant external pressure of 2 atm to a final pressure of 2 atm. During this process the temperature falls to -20 degree C. Calculate the heat, work, change in energy, and change in enthalpy for each step, as well the overall values (q_overall, W_overall, delta H__overall, delta H_ overall, ) as the gas transitions for the following paths. Note that for an ideal monoatomic gas: the molar heat capacity at constant volume (C^_v) is 3/2 R, and we will consider the heat capacities to be temperature independent. Reversible, isothermal expansion to the final volume followed by reversible, isochoric cooling to the final temperature. Reversible, isothermal expansion to the final pressure followed by reversible, isobaric cooling to the final temperature. Compare the values of q_ overall, w_ overall, delta E_ overall, and delta H_ overall, for the two paths. Comment on and justify any overall, overalls similarities or differences.

Explanation / Answer

Data Given:

P1=10 atm

T1=25 C=298.15 K

P2=2 atm

T2=-20 C=253.15 K

using ideal gas law

V1=nRT1/p1=1*8.314*298.15/10*101325=2.4464*10-3 m3

V2=8.314*253.15/2*101325=0.01 m3

(a)

step 1:isothermal expansion

So H=0

E=0

w=-nRTlnV2/V1 =-1*8.314*298.15*ln (0.01/2.4464*10-3)=-3490.09 J

q=U-w=3490.09 J

step 2;Reversible isochoric

So w=0

  U=CvT=3/2 R*(-20-25)=-3/2*8.314*45=-561 J

so For isochoric process q=U=-561 J

Cp-Cv=R

So Cp=5/2 R

  H=CpT=5/2*8.314*-45=-935.325 J

So overall

wov=-3490.09 J

qov=3490.09 J-561=2929.09 J

Uov=-561 J

Hov=-935.235 J

(b)

First step isothermal

H=0

U=0

w=-nRTln P1/P2=-8.314*298.15*ln 5=-3989.50 J

q=-w=3989.50 J

Step 2: isobaric process

H=CpT=5/2*8.314*-45=-935.325 J

q=-935.325 J

U=CvT=3/2*8.314*-45=-561 J

So

wov=-3989.50 J

qov=3989.50-935.325=3054.18 J

Uov=-561 J

Hov=-935.235 J

(c)

U and H values are same for both paths.Because they are state functions.

q and W are path function.So their values are different for both paths.

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