Entropy Change Confusion

  • #1
laser1
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Homework Statement
Calculate the change in entropy for the system when ##2## moles of a diatomic perfect gas, for which the molar heat capacity at constant pressure is ##\frac{7}{2} R##, is changed from 25 degrees Celsius and 1.50 atm to 135 degrees Celsius and 7.00 atm.
Relevant Equations
##\Delta S=\int_i^f \frac{dQ_\text{rev}}{T}##
1733586158472.png

Method 1 is the standard and (probably) the correct answer. I was fiddling around with it and don't understand why I get a different answer using method 2. If the gas is changing pressure, then it is changing volume. So why can't I sum up the entropy changes using method 2?

Using method 2 with V, by ##PV=nRT## I get ##V_i=32.61 \text{ L}## and ##V_f=9.33 \text{ L}##. So plugging in I end up with ##\Delta S = -7.7 \text{ J/K}##, whereas in method 1 I end up with ##-7.3 \text{ J/K}##. With method 2 with P I get ##-12.6 \text{ J/K}##.
 
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  • #2
Plot the paths on a P-V graph to see clearly what is happening.
Method 1: a. Heat from Ti to Tf at constant pressure Pi;
b. Compress from Pi to Pf at constant temp Tf.
Method 2: a. Heat from Ti to Tf at constant volume Vi;
b. Compress from Vi to Vf at constant temp Tf.
What is the pressure at the start of step 2b? It is not Pi.
Your mistake is in saying "at constant T, Vf/Vi = Pi/Pf".
For the whole process, T is not constant, so Vf/Vi ≠ Pi/Pf

By the way, for method 2 using V, I get the same as for method 1 using P (-7.3 J/K). Check your calculations.
 
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  • #3
And since you have the initial and final volumes, there's no reason to switch to pressures in method 2.
 
  • #4
mjc123 said:
Plot the paths on a P-V graph to see clearly what is happening.
Method 1: a. Heat from Ti to Tf at constant pressure Pi;
b. Compress from Pi to Pf at constant temp Tf.
Method 2: a. Heat from Ti to Tf at constant volume Vi;
b. Compress from Vi to Vf at constant temp Tf.
What is the pressure at the start of step 2b? It is not Pi.
Your mistake is in saying "at constant T, Vf/Vi = Pi/Pf".
For the whole process, T is not constant, so Vf/Vi ≠ Pi/Pf

By the way, for method 2 using V, I get the same as for method 1 using P (-7.3 J/K). Check your calculations.
Thank you. Drawing it out was a good idea. Yes, in method 2 I used temp of 125 degrees rather than 135 degrees! It all works out now, cheers.
 
  • #5
mjc123 said:
For the whole process, T is not constant, so Vf/Vi ≠ Pi/Pf
How do you derive ##\Delta S = nR\ln\left(\frac{P_i}{P_f}\right)## then?
 
  • #6
For method 1, step 1 is isobaric, so the whole change from Pi to Pf is isothermal (step 2).
For method 2 this is not true, so the equation does not apply.
 
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  • #7
mjc123 said:
For method 1, step 1 is isobaric, so the whole change from Pi to Pf is isothermal (step 2).
My understanding is that the formula ##\Delta S = nR\ln(V_f/V_i)## is still being used. However, the ##V_i## is not really the initial volume but the volume after the isobaric process is being done. Is this right?
 
  • #8
laser1 said:
My understanding is that the formula ##\Delta S = nR\ln(V_f/V_i)## is still being used. However, the ##V_i## is not really the initial volume but the volume after the isobaric process is being done. Is this right?
Yes, but I wouldn't say "really the initial volume." That expression is the entropy change of an ideal gas for an isothermal process. Given that context, the initial and final volumes are the volume at the start and end of the isothermal process. Neither is some random volume that happened to be given in a problem statement.
 
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