## Isothermal Irreversible

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

Abby Soriano 1J
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Joined: Sat Aug 24, 2019 12:16 am

### Isothermal Irreversible

Does anyone know how to calculate isothermal irreversible expansion? How does it differ from calculations for isothermal reversible expansion?

Ally Huang- 1F
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Joined: Thu Jul 25, 2019 12:16 am

### Re: Isothermal Irreversible

To calculate an isothermal irreversible expansion, use the equation W=-PDeltaV. Isothermal just indicates the process occurs at a constant temperature. For an isothermal reversible expansion, use the equation -nRTln(V2/V1).

Sartaj Bal 1J
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### Re: Isothermal Irreversible

A reversible process is one that can be reversed by an infinitely small change in a variable (infinitesimal). Usually the work a system can do is greatest in a reversible process. This is approximated as the maximum amount of work that can be performed.

Anika Chakrabarti 1A
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### Re: Isothermal Irreversible

I thought that the only way to have an isothermal irreversible expansion is if there is free expansion, meaning the external pressure is 0. Most irreversible expansions involve a temperature change.

Caitlyn Tran 2E
Posts: 100
Joined: Fri Aug 09, 2019 12:15 am

### Re: Isothermal Irreversible

Since both cases are isothermal, you can assume that delta U is 0. Work is calculated differently in these two scenarios because during an irreversible expansion, there is a large difference between the internal pressure and the external pressure, so expansion occurs quickly. This is why we can use $w = -P_{ext}\Delta V$ since there is a sudden change in volume. In a reversible expansion, the internal pressure is approximately equal to the external pressure, so change in volume occurs very slowly. This is why we integrate the usual equation we use to calculate work, which gives us $w = -nRTln(\frac{V_{2}}{V_{1}})$.

Skyllar Kuppinger 1F
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Joined: Thu Jul 25, 2019 12:16 am

### Re: Isothermal Irreversible

Caitlyn Tran 2E wrote:Since both cases are isothermal, you can assume that delta U is 0. Work is calculated differently in these two scenarios because during an irreversible expansion, there is a large difference between the internal pressure and the external pressure, so expansion occurs quickly. This is why we can use $w = -P_{ext}\Delta V$ since there is a sudden change in volume. In a reversible expansion, the internal pressure is approximately equal to the external pressure, so change in volume occurs very slowly. This is why we integrate the usual equation we use to calculate work, which gives us $w = -nRTln(\frac{V_{2}}{V_{1}})$.

but I remember there was a discussion worksheet where we were calculating the entropy change for an irreversible vs reversible system, and I believe my TA said that you CANNOT use -p delta V unless it is NON-isothermal. Did I misunderstand?

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

### Re: Isothermal Irreversible

Skyllar Kuppinger 1F wrote:
Caitlyn Tran 2E wrote:Since both cases are isothermal, you can assume that delta U is 0. Work is calculated differently in these two scenarios because during an irreversible expansion, there is a large difference between the internal pressure and the external pressure, so expansion occurs quickly. This is why we can use $w = -P_{ext}\Delta V$ since there is a sudden change in volume. In a reversible expansion, the internal pressure is approximately equal to the external pressure, so change in volume occurs very slowly. This is why we integrate the usual equation we use to calculate work, which gives us $w = -nRTln(\frac{V_{2}}{V_{1}})$.

but I remember there was a discussion worksheet where we were calculating the entropy change for an irreversible vs reversible system, and I believe my TA said that you CANNOT use -p delta V unless it is NON-isothermal. Did I misunderstand?

yes you cannot calculate -PdeltaV to find entropy and the equation nRln(\frac{V_{2}}{V_{1}}) must be used regardless of the process and that is because its a state function so the process that is taken does not matter. However, to find the work since it is not a state function, a distinction must be made and you have to use one of the two equations for work depending on the process taken since the process is taken into account.

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