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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?

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  1. Photon336
    • one year ago
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    \[pV = nRT \] 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 re-arrange 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

  2. Photon336
    • one year ago
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    @Zale101 did I do this right?

  3. anonymous
    • one year ago
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    nope

  4. Photon336
    • one year ago
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    wait, 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\[\frac{ -W }{ p } + V_{i} = V_{f}\]

  5. Photon336
    • one year ago
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    @aaronq

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