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anonymous
 one year ago
A sample of gas occupies a volume of 55.5 mL. As it expands, it does 140.6 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?
anonymous
 one year ago
A sample of gas occupies a volume of 55.5 mL. As it expands, it does 140.6 J of work on its surroundings at a constant pressure of 783 torr. What is the final volume of the gas?

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Photon336
 one year ago
Best ResponseYou've already chosen the best response.1\[pV = nRT \] pressure is constant. we need to figure out the definition of work \[work = pdV\] \[p \int\limits_{vi}^{vf} dv = p(v_{f}v_{i})\] \[p(V_{f}V_{i} = work \] Work is done on the surroundings so it's negative \[140.6J = p(V_{f}V_{I})\] we know the initial volume so we rearrange to find the final volume. J = joules. \[\frac{ work }{ pressure } = \frac{ J }{ p } + V_{i}\] 1.03 atm = 784 torr plug everything in \[\frac{ 140.6J }{ 1.03atm } +55mL =1.91 L\] @woodward @empty please check my math i'm bad at this stuff lol

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1@Zale101 did I do this right?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.1wait, I dont think it's negative pdV it's just pdV that would change the answer. 1. \[W = p(V_{f}V_{i})\] 2. \[\frac{ W }{ p } + V_{i} = V_{f}\] 3. My assumption was that work was done on the surroundings so Work was negative. 4. Which lead me to believe that the negatives canceled leaving us with\[\frac{ W }{ p } + V_{i} = V_{f}\]
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