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Introductory Physics Homework Help
Understanding the PV Diagram of a Monatomic Gas
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[QUOTE="PhizKid, post: 4388333, member: 402010"] [h2]Homework Statement [/h2] Consider the following PV diagram of one mole of a monatomic gas: [ATTACH=full]161546[/ATTACH] The units are in atm and litres. Find the change in internal energy, heat, and work for all 3 processes and for the cycle as well. [h2]Homework Equations[/h2] PV = nRT ΔU = Q - W ΔU = (3/2)*R*ΔT Q = (5/2)*R*ΔT (for constant pressure) [h2]The Attempt at a Solution[/h2] For the work, I just found the areas: A -> B: 0 J B -> C: +3.5 J C -> A: -1 J For the cycle, the work done is just the total work added: 3.5 - 1 = 2.5 J For the cycle, the Q is the same as the Work: 2.5 J For the cycle, the ΔU = 0. So far I have this (unless I did something wrong): [ATTACH=full]161547[/ATTACH] Since for a constant volume process the ΔU = Q, I decided to do that A -> B first. T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K T_B = (0.5 atm)(1 L) = (1)(0.0821)T, T = 36.5 K ΔT = 30.41 K ΔU = (3/2)*R*ΔT = (3/2)*0.0821*30.41 = 3.75 J = Q So: [ATTACH=full]161548[/ATTACH] The constant pressure process looks easier so I'll do C -> A next: T_C = (0.5 atm)(3 L) = (1)(0.0821)T, T = 18.27 K T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K ΔT = -12.18 Q = (5/2)*R*ΔT = (5/2)*0.0821*(-12.18) = -2.5 J ΔU = Q - W = (-2.5) - (-1) = -1.5 So: [ATTACH=full]161549[/ATTACH] Now I can just add up the unknown columns: ΔU for B -> C: 3.75 + (-1.5) + x = 0, x = -2.25 Q for B -> C: 3.75 + (-2.5) + x = 2.5, x = 1.25 So I got: [ATTACH=full]161550[/ATTACH] But I got this problem incorrect. What errors have I made? [/QUOTE]
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Introductory Physics Homework Help
Understanding the PV Diagram of a Monatomic Gas
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