Isothermal, isovolumetric, and isobaric processes in ideal gases

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SUMMARY

This discussion focuses on the analysis of isothermal, isovolumetric, and isobaric processes involving one mole of an ideal gas in a cylinder with a movable piston. The work done on the gas during isobaric compression is calculated as P(1/2)V, while for isothermal compression, the work is expressed as W = -nRTln(1/2)P. The isovolumetric process results in zero work done (W = 0). Participants also matched each process to its operational description, confirming that isothermal corresponds to B, isovolumetric to C, and isobaric to A.

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Homework Statement



One mole of an ideal gas is contained in a cylinder with a movable piston. The initial pressure, volume, and temperature are P, V, and T, respectively. Consider each of the following processes:
· isobaric compression in which the final volume is one-half the initial volume
· isothermal compression in which the final pressure is two times the initial pressure
· isovolumetric process in which the final pressure is two times the initial pressure

Determine the work done on the gas for each process. (Use any variable or symbol stated above as necessary.):

isobaric compression:

Isothermal Compression:

Isovolumetric compression:



in In operational terms, match each process with its description of how to carry out the process.
isothermal
isovolumetric
isobaric

A. Put the cylinder into a refrigerator at absolute temperature T/2. Let the piston move freely as the gas cools.

B. With the gas in a constant-temperature bath at T, slowly push the piston in.

C. Lock the piston in place and hold the cylinder over a hotplate at 2T.

Homework Equations





The Attempt at a Solution



For Isobaric compression:

Constant pressure , also gas is being compressed so W is positive

so P(1/2)V... since Vf is half Vi

for Isothermal compression:

gas is not under Cp. so evaluate the integral dW=-Pdv

W=-nRTln(1/2)P

and finally

Isovolumetric compression: w=0.

for part b matching the process to the right statement.

Isothermal:B

IsoVolumetric:C

IsoBaric: A

is this correct?
 
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All your approaches are correct :smile: Just a minute error, I found..

charmedbeauty said:
gas is not under Cp. so evaluate the integral dW=-Pdv

W=-nRTln(1/2)P

The initial pressure is P, the final pressure is 2P. So what is ln(\frac{P}{2P})? :wink:
 
Infinitum said:
All your approaches are correct :smile: Just a minute error, I found..



The initial pressure is P, the final pressure is 2P. So what is ln(\frac{P}{2P})? :wink:

oops should be ln(1/2)
 
charmedbeauty said:
oops should be ln(1/2)

are you sure this is right I put it in as nRTln(1/2) and it said it was wrong??
 
charmedbeauty said:
are you sure this is right I put it in as nRTln(1/2) and it said it was wrong??

You seem to be forgetting a negative sign.
 
Infinitum said:
You seem to be forgetting a negative sign.

oops, thanks infinitum
 
charmedbeauty said:
oops, thanks infinitum

nope still wrong.
 
charmedbeauty said:
nope still wrong.

:confused:

It seems correct to me, maybe you need to give the value of ln(1/2) or base change it to 10 instead?

Edit : I just re-read the question...

One mole of an ideal gas is contained in a cylinder with a movable piston.

Doh!
 
Infinitum said:
:confused:

It seems correct to me, maybe you need to give the value of ln(1/2) or base change it to 10 instead?

Edit : I just re-read the question...



Doh!

oh so just get rid of n
 
  • #10
charmedbeauty said:
oh so just get rid of n

nope still wrong.
 
  • #11
charmedbeauty said:
nope still wrong.

This most definitely has to be the correct answer. Assuming you didn't forget the negative sign, I can't see anything wrong with it.

Did you try out putting in the value of ln(1/2) or changing the base to 10?
 

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