Entropy Change - Irreversible Isothermal Compression

In summary, the conversation discusses calculating the entropy change of a system undergoing an isothermal irreversible compression, with a constant external pressure of 10 atm. The change in internal energy is assumed to be zero, and the work done on the system is equal to the heat removed from the system. It is also mentioned that the sign convention for heat transfer can be fuzzy, but it is accurate to say that heat removed from the system is equivalent to heat gained by the surroundings.
  • #1
Stephen Clarke
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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.
 
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  • #2
How much heat is transferred to the surroundings (which can be considered as an ideal reservoir at 253 K)?
 
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  • #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
Stephen Clarke said:
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.
 
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  • #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
Stephen Clarke said:
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)?
Stephen Clarke said:
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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Related to Entropy Change - Irreversible Isothermal Compression

1. What is entropy change?

Entropy change is a measure of the increase or decrease in the disorder or randomness of a system as it undergoes a process.

2. How is entropy change calculated?

In the case of irreversible isothermal compression, entropy change can be calculated using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.

3. What is irreversible isothermal compression?

Irreversible isothermal compression is a process in which a gas is compressed at a constant temperature, but in a way that is not reversible. This means that the system cannot be returned to its original state without energy input from an external source.

4. Why is irreversible isothermal compression important?

Irreversible isothermal compression is important because it occurs in many practical processes, such as the compression of gases in engines and refrigerators. It also helps us understand the relationship between entropy change, heat transfer, and temperature.

5. How does entropy change during irreversible isothermal compression affect the efficiency of a process?

The increase in entropy during irreversible isothermal compression reduces the efficiency of a process, as some of the energy is lost as heat. This decrease in efficiency is known as the Carnot efficiency, and it is a fundamental limit for the efficiency of any heat engine operating between two temperatures.

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