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Entropy Change - Irreversible Isothermal Compression

  • #1
Given a sample of nitrogen gas (assume ideal gas conditions), the following conditions were observed inside the container. n = 0.75 moles at 253 K, and pressure = 0.5 atm. Then, an ISOTHERMAL IRREVERSIBLE COMPRESSION on the system forced by a constant Pexternal = 10 atm reduced the initial volume by a factor of 2. Assume diathermal walls on this container. Calculate the entropy change of the system, surroundings and universe.

ATTEMPT:

The entropy change of the system is under ISOTHERMAL conditions, so, dT = 0, and,

dS = nRln(P2/P1)
dS = (0.75 mol)(8.314 J/Kmol)ln(1/2)

However, I am having a difficult time conceptualizing the entropy change of the surroundings. This is obviously an integral part of computing the entropy change of this particular universe.

I'll appreciate any feedback! Thank you.
 

Answers and Replies

  • #2
19,917
4,094
How much heat is transferred to the surroundings (which can be considered as an ideal reservoir at 253 K)?
 
  • #3
AH! To answer your question, is it correct to say that the heat transferred to the surroundings is simply q = -w?

That is to say, work done by the constant external Pressure forces heat out of the system, or w = -P(V2-V1)?
 
  • #4
19,917
4,094
AH! To answer your question, is it correct to say that the heat transferred to the surroundings is simply q = -w?
Is the change in internal energy zero?
I assume you are using the form of the first law in which work done on the system is positive and work done by the system is negative: ##\Delta U=q+w##. Correct?
That is to say, work done by the constant external Pressure forces heat out of the system, or w = -P(V2-V1)?
I wouldn't say it in those words. Instead, I would say that, since the change in internal energy is zero, the work done on the system is equal to the heat removed from the system. Your equation for the work is correct, if work done on the system is taken as positive.
 
  • #5
Thanks. One last question.

I know sign convention and interpretation can get fuzzy. Is it accurate to say heat removed from this system (-Q) is equivalent to heat gained by the surroundings, in which case we would use a positive value of Q? (Since we are changing reference points, I.e., system --> surroundings.)
 
  • #6
19,917
4,094
AH! To answer your question, is it correct to say that the heat transferred to the surroundings is simply q = -w?

That is to say, work done by the constant external Pressure forces heat out of the system, or w = -P(V2-V1)?
Thanks. One last question.

I know sign convention and interpretation can get fuzzy. Is it accurate to say heat removed from this system (-Q) is equivalent to heat gained by the surroundings, in which case we would use a positive value of Q? (Since we are changing reference points, I.e., system --> surroundings.)
Sure.
 

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