Showing work done on gas in a reversible process

In summary, the problem involves an ideal gas initially at temperature Ti, pressure Pi, and volume Vi being compressed reversibly down to half its original volume. The temperature of the gas varies during the compression to maintain the equation P=AV. The goal is to show that the work done on the gas is -\frac{3}{8}nRTi. To solve this, we use the equation W=\int PdV and substitute P=AV. From there, we can determine the value of A using the ideal gas equation.
  • #1
Ryomega
67
0

Homework Statement



Ideal gas initially at temperature Ti, Pressure Pi, Volume Vi is compressed reversibly down to half its original volume. Temperature of gas is varied during the compression so that

P = AV is always satisfied [where A is a constant]

Show that the work done on the gas is:

-[itex]\frac{3}{8}[/itex]nRTi

Homework Equations



W = ∫ Pdv

W = nRT ln([itex]\frac{Vf}{Vi}[/itex])

W = A[itex]\frac{Vf-Vi}{1-\gamma}[/itex]

[itex]\gamma[/itex] = [itex]\frac{Cp}{Cv}[/itex]=[itex]\frac{N+2}{N}[/itex] where N is the degree of freedom (3 in this case)

The Attempt at a Solution



I gather that the solution is already in one of those equations and that I am probably being stupid. Yet everything I have tried so far does not get me to the solution.

From the problem I understand that the process is adiabatic since process is reversible and temperature changes to give:

[itex]\frac{P}{V}[/itex]=const.

which I am confused on since:

PV^[itex]\gamma[/itex]=const

I understand that this may be against the policy of this forum, but would anyone mind showing me how to arrive at this solution? I learn best from examples.

Thank you
 
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  • #2
Hi Ryomega. The work done W=nRTln([itex]\frac{Vf}{Vi}[/itex]) doesn't apply here because the temperature isn't constant. So from W=∫PdV, and use P=AV in the integral from the initial volume of V to the final volume of V/2. You can workout what A is from the ideal gas equation.
 
  • #3
Aha! I knew I was being stupid! I didn't think about the substitution. Thanks a LOT!
 

Related to Showing work done on gas in a reversible process

1. What is a reversible process in terms of gas?

A reversible process in gas refers to a process in which the system changes from one state to another and then back to its original state without any changes in the surroundings. This means that the process can be reversed without any losses or changes in energy.

2. Why is it important to show work done on gas in a reversible process?

Showing work done on gas in a reversible process allows us to calculate the amount of energy that is converted into work. This is important because it helps us understand the efficiency of the process and how much energy is lost or gained during the process.

3. How do you calculate work done in a reversible process?

The work done in a reversible process can be calculated using the equation W = P∆V, where P is the pressure and ∆V is the change in volume. This equation assumes that the process is isothermal, meaning that the temperature remains constant.

4. Can you give an example of a reversible process in gas?

One example of a reversible process in gas is an isothermal expansion or compression. In this process, the gas expands or compresses at a constant temperature, meaning that the process is reversible and no energy is lost.

5. How does showing work done on gas in a reversible process relate to the laws of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only converted from one form to another. Showing work done on gas in a reversible process allows us to see how energy is converted into work, which is in accordance with the first law of thermodynamics.

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