How do you calculate isothermal expansion?

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How do you calculate isothermal expansion?

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Answer 1 · 2017-07-04T19:20:43

[I do it at the second half of this answer here.](https://socratic.org/questions/what-is-reversible-isothermal-expansion)

Copy-and-pasted, we get...

CALCULATION EXAMPLE

Calculate the work performed in a reversible isothermal expansion by $1$ $1$ $m o l$ $mol$ of an ideal gas from $22.7$ $22.7$ $L$ $L$ to $45.4$ $45.4$ $L$ $L$ at $298.15$ $298.15$ $K$ $K$ and $1$ $1$ $b a r$ $ba r$ .

With the ideal gas law, we have that $P V = n R T$ $PV = nRT$ , or $P = \frac{n R T}{V}$ $P = (nRT)/V$ . So, the work is:

$w_{r e v} = - \int_{V_{1}}^{V_{2}} P d V$ $color(green)(w_(rev)) = -int_(V_1)^(V_2) PdV$

$= - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V$ $= -int_(V_1)^(V_2) (nRT)/VdV$

$= - n R T {ln V}_{2} - (- n R T {ln V}_{1})$ $= -nRTlnV_2 - (-nRTlnV_1)$

$= - n R T ln (\frac{V_{2}}{V_{1}})$ $= color(green)(-nRTln(V_2/V_1))$ ,

negative with respect to the system.

We keep in mind that the pressure did change, but we don't have an idea of how, off-hand. The work thus does not use the pressure of $1 bar$ $"1 bar"$ :

$w_{r e v} = - (1 mol) (8.314472 J/mol \cdot K) (298.15 K) ln (\frac{45.4 L}{22.7 L})$ $color(blue)(w_(rev)) = -("1 mol")("8.314472 J/mol"cdot"K")("298.15 K")ln("45.4 L"/"22.7 L")$

$=$ $=$ $- 1718.3 J$ $color(blue)(-"1718.3 J")$

So, the work involved the ideal gas exerting $1718.3 J$ $"1718.3 J"$ of energy to expand, expunging $1718.3 J$ $"1718.3 J"$ of heat from itself:

$cancel(DeltaU)^(0" for isothermal process") = q_(rev) + w_(rev)$

$=> color(blue)(q_(rev)) = -w_(rev) = color(blue)(+"1718.3 J")$

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How do you calculate isothermal expansion?

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