Work Done by Isothermal Expansion

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SUMMARY

The discussion focuses on calculating the work done during the isothermal expansion of an ideal gas (0.800 moles) from a pressure of 10.0 atm to 1.0 atm at a temperature of 200K. The participants clarify that work can be calculated using the formula dW = ∫ PdV, with specific cases for expansion against a vacuum, against a constant external pressure of 1.0 atm, and reversibly. For the reversible case, the work done is expressed as W = nRT ln(V2/V1), while for the constant external pressure, it is the product of the external pressure and the change in volume. In the case of expansion against a vacuum, no work is done.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV = nRT)
  • Knowledge of thermodynamic work calculations
  • Familiarity with concepts of isothermal processes
  • Basic integration techniques for calculating work
NEXT STEPS
  • Learn how to derive work done in isothermal processes using the Ideal Gas Law
  • Study the differences between reversible and irreversible processes in thermodynamics
  • Explore the concept of free expansion and its implications on work done
  • Investigate the relationship between pressure, volume, and work in various thermodynamic cycles
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Students studying thermodynamics, physics enthusiasts, and professionals in engineering fields who require a solid understanding of gas behavior and work calculations in isothermal processes.

Calu
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Homework Statement


A quantity of ideal gas (0.800mole) at a pressure of 10.0atm and 200K is allowed to
expand isothermally until it reaches a pressure of 1.00atm. Calculate the work done
if this expansion is carried out a) against a vacuum, b) against a constant external
pressure of 1.0atm and c) reversibly.

I'm not even sure where to start here. I've only calculated work done using dW = ∫ PdV. However, I'm not given an initial or final volume here, so I'm not sure how to proceed. Any hints?
 
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It is ideal gas. Given the amount of gas, the pressure and the temperature. What is the volume?
 
ehild said:
It is ideal gas. Given the amount of gas, the pressure and the temperature. What is the volume?
I see, I simply rearrange PV = nRT?

In which case, how does my calculation change between each of the above instances?
 
Calu said:
I see, I simply rearrange PV = nRT?

In which case, how does my calculation change between each of the above instances?
I do not see any calculations of yours. :)
 
ehild said:
I do not see any calculations of yours. :)

How would I go about calculating work when both the pressure and the volume change? I can integrate P.dV and put in the values for the initial and final volumes, but what value of pressure do I use?
 
Calu said:
How would I go about calculating work when both the pressure and the volume change? I can integrate P.dV and put in the values for the initial and final volumes, but what value of pressure do I use?
Pressure is only defined for reversible (quasi-static) process. In that case, you can express P from the ideal gas law.

In the other cases, you can calculate the work done on the gas by external forces.
What is the work done on the gas if there is no external force (when the gas expands into vacuum)?
What is the work done on the gas if the external pressure is constant (1.0 atm)?
 
ehild said:
Pressure is only defined for reversible (quasi-static) process. In that case, you can express P from the ideal gas law.

In the other cases, you can calculate the work done on the gas by external forces.
What is the work done on the gas if there is no external force (when the gas expands into vacuum)?
What is the work done on the gas if the external pressure is constant (1.0 atm)?

Okay, so in the reversible case I can substitute nRT ∫ 1/V .dV = nRT ln(V2-V1). I'm not sure how to work out the other two cases though.
 
What is the work done if no force is exerted?
What is the work done by the environment on the gas if the external pressure is 1.0 atm and the volume changes from V1 to V2?
 
ehild said:
What is the work done if no force is exerted?
What is the work done by the environment on the gas if the external pressure is 1.0 atm and the volume changes from V1 to V2?

Okay, so in the first case, I would say no work is being done on the gas, however the question just asks for the work done. Is the gas doing work on the surroundings in this case? In the second case I would say that the numerical value of work done is equivalent to the difference between the volumes.
 
  • #10
Calu said:
Okay, so in the first case, I would say no work is being done on the gas, however the question just asks for the work done. Is the gas doing work on the surroundings in this case?
Is there anything to do work on if the gas expands freely?
Calu said:
In the second case I would say that the numerical value of work done is equivalent to the difference between the volumes.
It is the constant external pressure multiplied by the difference of volumes, and the external pressure is 1.0 atm,
 
Last edited:
  • #11
ehild said:
Is there anything to do work on if the gas expands freely?

It is the constant external pressure multiplied by the difference of volumes, and the external pressure is 1.0 atm, so you are right...

In first case, I guess not. That makes sense. In the second case I apologise for omitting the multiplication by 1, its how I've been taught. I very really show multiplication by 1 in my working.
 
  • #12
Calu said:
In first case, I guess not. That makes sense. In the second case I apologise for omitting the multiplication by 1, its how I've been taught. I very really show multiplication by 1 in my working.
It is not multiplication by 1 but multiplication by 1.0 atm. And the result should be in joules. So what is the numerical value at the end?
 

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