New Notes Blog: Find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. A quantity of...

Tuesday, 24 September 2013

Find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. A quantity of...

Work is done when a constant force F is applied to move an object a distance D. It is defined with a formula $\displaystyle{W}={F}{D}.$

For expanding gas, we denote the work done as $\displaystyle{W}={P}\cdot\delta{V}.$

With the stated assumption pressure is inversely proportional to volume, we let $\displaystyle{P}=\frac{{k}}{{V}}$ where k is the proportionality constant.

Then plug-in $\displaystyle{P}=\frac{{k}}{{V}}$ on $\displaystyle{W}={P}\cdot\delta{V}$ , we get: $\displaystyle{W}=\frac{{k}}{{V}}\cdot\delta{V}$ ...

Work is done when a constant force F is applied to move an object a distance D. It is defined with a formula $\displaystyle{W}={F}{D}.$

For expanding gas, we denote the work done as $\displaystyle{W}={P}\cdot\delta{V}.$

With the stated assumption pressure is inversely proportional to volume, we let $\displaystyle{P}=\frac{{k}}{{V}}$ where k is the proportionality constant.

Then plug-in $\displaystyle{P}=\frac{{k}}{{V}}$ on $\displaystyle{W}={P}\cdot\delta{V}$ , we get: $\displaystyle{W}=\frac{{k}}{{V}}\cdot\delta{V}$ or $\displaystyle\frac{{{k}\delta{V}}}{{V}}$

The integral of work done will be $\displaystyle{W}={\int_{{{V}_{{1}}}}^{{{V}_{{2}}}}}\frac{{{k}{d}{V}}}{{V}}$

To solve for the proportionality constant $\displaystyle{\left({k}\right)}$ , we plug-in the initial condition:

$\displaystyle{P}={2500}$ pounds and $\displaystyle{V}_{{1}}={1}$ on $\displaystyle{P}=\frac{{k}}{{V}}.$

$\displaystyle{2500}=\frac{{k}}{{1}}$

$\displaystyle{k}={2500}\cdot{1}={2500}$

To solve for the work done by the gas to expand the volume, we plug-in $\displaystyle{k}={2500}$ , $\displaystyle{V}_{{1}}={1}$ , and $\displaystyle{V}_{{2}}={3}$ on $\displaystyle{W}={\int_{{{V}_{{1}}}}^{{{V}_{{2}}}}}\frac{{{k}{d}{V}}}{{V}}$ .

$\displaystyle{W}={\int_{{{1}}}^{{{3}}}}\frac{{{2500}{\left({d}{V}\right)}}}{{V}}$

Apply basic integration property: $\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$ .

$\displaystyle{W}={2500}{\int_{{{1}}}^{{{3}}}}\frac{{{d}{V}}}{{V}}$ .

Apply basis integration formula for logarithm.

$\displaystyle{W}={2500}{\ln{{\left({V}\right)}}}{{\mid}_{{{1}}}^{{{3}}}}$

Apply definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{W}={2500}{\ln{{\left({3}\right)}}}-{2500}{\ln{{\left({1}\right)}}}$

$\displaystyle{W}={2500}{\left[{\ln{{\left({3}\right)}}}-{\ln{{\left({1}\right)}}}\right]}$

Apply natural logarithm property: $\displaystyle{\ln{{\left(\frac{{x}}{{y}}\right)}}}={\ln{{\left({x}\right)}}}-{\ln{{\left({y}\right)}}}$ .

$\displaystyle{W}={2500}{\left[{\ln{{\left(\frac{{3}}{{1}}\right)}}}\right]}$

$\displaystyle{W}={2500}{\ln{{\left({3}\right)}}}$ or $\displaystyle{2746.53}$ ft-lbs.

New Notes Blog

Tuesday, 24 September 2013

Find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. A quantity of...

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Tuesday, 24 September 2013

Find the work done by the gas for the given volume and pressure. Assume that the pressure is inversely proportional to the volume. A quantity of...

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?