Chapter 8 Question 11 Part B  [ENDORSED]

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Raymond_Guan_1H
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Chapter 8 Question 11 Part B

Postby Raymond_Guan_1H » Thu Jan 19, 2017 11:52 pm

How would you know to use the w(variable pressure) equation when the information is only stating it's expanding reversibly and isothermally?

Marie_Bae_3M
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Joined: Wed Sep 21, 2016 2:57 pm

Re: Chapter 8 Question 11 Part B  [ENDORSED]

Postby Marie_Bae_3M » Fri Jan 20, 2017 2:00 am

I think that "isothermal" (constant temperature) "reversible" expansion was meant to emphasize a certain work equation we have to use: w = -nRTln(Vfinal/Vinitial)?
"Isothermal" in that as a gas expands (increase in volume), the gas's pressure decreases, as stated by Boyle's law (P1V1=P2V2, constant temperature). This is crucial because in order to achieve reversible expansion, the external pressure has to change in step with the change in volume to keep the pressure of the gas and the external pressure same.
Since a reversible process is one that can be reversed by an "infinitely small change in a variable," this means we have to use calculus to get the derivative of the function for work. So, dw=-P(external)*dv or dw=-P*dv (because P(external)=P) or dw=(nRT/V)*dv (because PV=nRT). The total work done is given by taking the integral of all of these "infinitely small" changes, therefore taking the integral of dw. -nRT are constants because it is at constant temperature, and solving the integral gives us w=-nRTln(Vfinal/Vinitial). We use this to find the w of the reversible expansion: w = -(0.200mol)(8.314J/(K*mol))(298.15K)(ln(2.40L/1.20L). We know Vfinal = 2.40 because a states that the gas expands an additional 1.20 atm from the original 1.20 atm.


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