Question

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?

Answer 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 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

Answer 2

@Zale101 did I do this right?

Answer 3

nope

Answer 4

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\[\frac{ -W }{ p } + V_{i} = V_{f}\]

Answer 5

@aaronq