Expansion of an Ideal gas(molecular thermodyanmics)

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The discussion focuses on calculating thermodynamic properties for one mole of a monoatomic ideal gas transitioning from a pressure of 2.00 bar to 4.00 bar along a reversible path defined by P/V = constant. The user has successfully calculated the change in internal energy (Delta(U) = 10.23 KJ*mol^-1) and change in enthalpy (Delta(H) = 17.01 KJ*mol^-1) but is struggling with the work calculation due to the unique path condition. A suggestion is made to plot the process on a PV diagram to visualize the linear relationship between pressure and volume, which aids in understanding the work done during the expansion. The user is advised to substitute the relationship of pressure in the work integral to find the solution. The discussion emphasizes the importance of understanding the path taken in thermodynamic processes to accurately compute work.
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I'm having trouble with harder part of this question.

Homework Statement


One mole of monoatomic ideal gas initially at a pressure of 2.00 bar and temperature of 273 K is taken to a final pressure of 4.00 bar by the reversible path defined by P/V = constant. Calculate the values of Delta(H)(change in enthalpy), Delta(U)(potential energy), q(heat) and w(work). Cv = 12.5 kj*mol^-1*K^-1


Homework Equations


Delta(U) = q + w
Delta(U)= Cv*delta(t)
Delta(H)= Cp*delta(t)
Cv = 12.5 kj*mol^-1*K^-1
Cp = Cv - R
w = -Pext*delta(V)
q = Delta(U) - w
R = .08314 dm^3*bar*mol^-1*K^-1

I'm missing the equation for work in this situation as the reversible path is defined by P/V = constant.


The Attempt at a Solution



I found the change in enthalpy and the potential energy. They are correct, the textbook lists quantitative answers. However, the work is proving difficult as I don't understand the reversible path condition(P/V = constant) and how to relate that the energy of work. I know that it relates in some way to the ideal gas equation of state, but I cannot figure out that way. Thank you, I'm not sure if this should be introductory physics or advanced forum, so I've placed it in both.

What I've calculated so far, all are correct according to the text:

Delta(U) = 10.23 KJ*mol^-1
Delta(H) = 17.01 KJ*mol^-1

V1 = 11.35 dm^3
V2 = 22.7 dm^3
T1 = 273 K
T2 = 1092 K

thanks again
 
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Hint: plot that reversible path on the P(V) diagram. ehild
 
Okay so if i plot this in PV diagram, i see a linear relationship where both P and V get larger wrt each other where the linear fit is P=2V. Alright, now this isn't an isothermal process. I can't see how to relate work without that condition. I know there is PdV, but i don't see the next step i guess.
 
casiobeats said:
Okay so if i plot this in PV diagram, i see a linear relationship where both P and V get larger wrt each other where the linear fit is P=2V.

P=const * V, but the constant is not 2. Find it. You know the pressure and temperature at the initial state, and have calculated that the volume is 11.35 dm^3. What is P/V?

The work is

W = \int_{V1}^{V2}{PdV}

P=const*V. Substitute this for P in the integrand.

ehild
 
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