Isobaric Process: Finding Q, E, W.

AI Thread Summary
The discussion focuses on calculating the change in energy, heat (Q), and work (W) for an isobaric process involving 1.46 moles of an ideal gas heated from 431°C to 1227°C. The change in internal energy (IE) is calculated using the formula for ideal gases, resulting in 14.5 kJ. The challenge arises in determining the initial and final volumes to calculate work, as both pressure and volume are unknowns. Participants suggest using the ideal gas law to relate pressure and volume changes to temperature changes, emphasizing that only the combination PΔV is needed. Additionally, caution is advised regarding the signs in the equation ΔIE = Q + W, clarifying the distinction between work done on the system versus work done by the system.
yaylee
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Homework Statement



n = 1.46 moles of ideal gas are heated isobarically (at constant pressure) from temperature To = 431 oC to temperature Tf = 1227 oC. Find: change in Energy, Q, W.

Homework Equations



Change in IE = Q + W
W (isobaric process) = P(vf-v1)

The Attempt at a Solution


change in IE = change in KE (because it is an ideal gas) = (3/2)nRT.
So, (3/2)(1.46)(8.314 x 10^-3)(1227-431) = 14.5 kJ (no issue here)

W = P(Vf-Vi), since it is an isobaric process.
However, how are you supposed to get the volumes of the initial and final state?

I tried pv = nRT, but am stuck with two unknowns. (Both Pressure, and Volume). Any suggestions? Many thanks in advance !
 
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Hint: You don't need separate values for P and V. You only need the combination PΔV.
Can you relate PΔV to ΔT using the ideal gas law?

Also, be careful with signs. You wrote ΔIE = Q + W. So, W is work done on the system. Does PΔV represent work done on the system or work done by the system?
 

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