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PV diagram of a monatomic ideal gas

  • Thread starter mikefitz
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  • #1
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Homework Statement





Suppose 0.50 mol of a monoatomic ideal gas is changed from state A to state D by one of the processes shown on the PV diagram of the figure below.

http://img382.imageshack.us/img382/784/untitledsc4.png [Broken]

(This is my edited version of the image)

(a) Find the total work done by the gas if it follows the constant pressure path A-E followed by the constant temperature path E-D.

(b) Calculate the total change in internal energy of the gas during the entire process and the total heat flow into the gas.




I'm having trouble with (a). Is the work done from path A-E, E-D reflected on the first diagram I made, or the second? I guess I'm confused as to whether I'm supposed to use 1 or 2 atm in my calculations. thanks
 
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Answers and Replies

  • #2
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Hi Mikefitz,

It has been a while since i touched my thermodynamics but nevertheless i'll try to help.
For part (a),
U need to know that 1 atm = 1.01 x 10^5 Pa and 1 litre = 1000 cm³ = 0.001 m³
Work done from A-E (isobaric) = P dV = 2 x 1.01 x 10^5 Pa x (8-4) x 0.001m³ = 808 J
Work done from E-D (isothermal) = Area of trapezium E-D = 1/2 (2+1) x 1.01 x 10^5 Pa x (16-8) x 0.001m³ = 1212 J
Total work done by gas (expansion) = 808 J + 1212J = 2020 J

For part (b),
U use the 1st law of thermodynamics, dU = Q + W or dU = 3/2nRT to solve for the change in internal energy, dU and heat flow, Q, into the gas.

Hope my A level physics helps. Correct me if i'm wrong.
 

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