GREPhysics.NET: GR0177 #36

GR0177 #36

Problem

GREPhysics.NET Official Solution

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Thermodynamics $\Rightarrow$ }Adiabatic

In an adiabatic expansion, $dS=0$ (entropy), since $dQ=0$ (heat). No heat flows out, by definition, and thus choice (A) is out, as well as choice (B). Choice (C) is true since by the first law, one has $Q=U+W \Rightarrow U=-W$ , and the given integral is just the definition of work. Choice (D) defines work. Choice (E) remains---so take that.

Alternate Solutions

Ning Bao
2008-02-01 06:29:07

You could also remember that adiabatic and isothermal are not synonyms -> E.

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Comments

anmuhich 2009-03-14 12:57:30	I agree with Ning Bao. It's easiest just to realize that in an adiabatic process the temperature changes, otherwise it would be isothermal. Reply to this comment
phys2718 2008-09-24 15:58:27	Careful, zero heat change alone isn't enough to guarantee zero entropy change. For example, in the adiabatic free expansion of the ideal gas there is an increase in entropy. The other necessary condition for no change in entropy is that the process be reversible, which is true here since the problem stated that we have a quasi-static expansion rather than a free expansion Reply to this comment
Ning Bao 2008-02-01 06:29:07	You could also remember that adiabatic and isothermal are not synonyms -> E. Reply to this comment
antithesis 2007-10-01 12:19:36	As a shorter way to do this exists if you remember that for adiabatic process, $PV^\gamma = constant$ or alternativly, $TV^{\gamma-1} = constant$ Therefor, if V changes ( $V_i \neq V_f$ ), T must change, and so (E) can't be true. Reply to this comment

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I agree with Ning Bao. It's easiest just to realize that in an adiabatic process the temperature changes, otherwise it would be isothermal.

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LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
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$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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