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I am given the information that a parcel of air expands adiabatically (no exchange of heat between parcel and its surroundings) to five times its original volume, and its initial temperature is 20° C. Using this information, how can I determine the amount of work done by the parcel on its surroundings?

I know that $dq = 0$, and that $du + dW = dq = 0$, but I don't know what to do with this information. $dW = pdV$, which seems like it should be helpful, but I don't know what to do for the pressure.

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  • $\begingroup$ I just wanted to comment because some people are bound to think this violates our homework policy - but personally I think you just make it into the domain of acceptable homework-like questions because you narrowed the problem down to the underlying concept, which is that you don't see a way to get the pressure. $\endgroup$
    – David Z
    Commented Oct 9, 2012 at 17:58

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More clues? :-)

This is harder then the isobaric process because now the pressure is a function of volume. You need to write the pressure as a function of volume, then integrate it from the initial to final volume. For some clues see the Wikipedia article on adiabatic expansion. Although the question doesn't say so, you'll need to assume the expansion is reversible as the question can't be answered otherwise.

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  • $\begingroup$ If I use $pV = RT$, then $p = \frac{RT}{V}$. Using $\delta w = pdV$, then $w = RT \int{\frac{1}{V}} dV$. Solving this gives me $w = RT\ln(5)$, which, after plugging numbers in, gives me $w = (287)(293)\ln5 = 1.35339 * 10^5$. And I was just sent an email saying that the book's answer is wrong and this one is correct. Thank you so much for your help, I really appreciate it (and I do completely get it now)! $\endgroup$
    – vaindil
    Commented Oct 9, 2012 at 18:42
  • $\begingroup$ Hmm, actually that's wrong. The temperature is a function of volume as well, so you can't just assume the pressure is constant. The way to do it is to note that for a reversible adiabatic expansion $PV^\gamma$ is a constant, where $\gamma$ is a constant related to the type of gas. $\endgroup$
    – John Rennie
    Commented Oct 9, 2012 at 19:31
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$PV^\gamma=\rm constant$ for adiabats,so use $$ W=\int\frac{\rm const}{V^γ} dV $$ then recall that the constant in the integration is $PV^\gamma$. You should get $$ \frac{1}{(1-\gamma)}(PV) $$ evaluated between the two volumes and pressures.

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dq=0 in an adiabatic reversible process. From First Law of Thermodynamics,

dE = dq + dW (But dq=0)

Therefore dE = dW. It implies that

nCvdT = -Pdv

Cv = (-RT/V)dv

Cv(dt)/T = -Rdv/V

On integrating with limits T1,T2 and V1,V2,

(T2/T1)Cv/R = V1/V2.

( I will post the rest of the answer later ..........)

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    $\begingroup$ Please format your equations using MathJax syntax $\endgroup$
    – nluigi
    Commented Oct 7, 2015 at 10:14

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