## Reversible and Isothermal

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Yen Vuong 2I
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### Reversible and Isothermal

How does a reversible piston and an isothermal system affect work?

Courteney Hedicke 3J
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### Re: Reversible and Isothermal

I think Isothermal means that the temperature doesnt change.

TiengTum2D
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### Re: Reversible and Isothermal

I think isothermal system is a system that has a consistent heat source around it and it can affect work by either expansion or compression with the piston. If the gas expands (piston moves out), the system is doing work but heat around the area will replenish the lost energy as heat, therefore keeping it at the same temperature. I hope that makes sense!

Coco Hailey 2E
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### Re: Reversible and Isothermal

When calculating the work of a reversible or isothermal system, do we use different equations. Basically if its a reversible system do we use w= -P∆V or w = -nRTln(V2/V1)??

Pauline Tze 3B
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### Re: Reversible and Isothermal

Hi Coco!

A reversible system is typically isothermal because the heat absorbed by the system is canceled out by the heat released by the system (as is the nature of systems at equilibrium). We use those two specific equations you listed as follows:
w= -P∆V is to find work in a system where there's constant pressure
w = -nRTln(V2/V1) is to find work when there's variable pressure (or if the system is, for example, expanding/increasing in volume). Of the two equations, w = -nRTln(V2/V1) is appropriate for reversible reactions because the external pressure changes forcing the system to "reverse" that action by moving toward equilibrium.

I hope that helps!