# How do you find heat in isothermal processes?

Jun 30, 2017

For ideal gases, which are usually what you'll deal with in calculations involving isothermal processes, the internal energy is a function of only temperature.

That means the first law of thermodynamics becomes:

cancel(underbrace(DeltaU)_"change in internal energy")^(0) = underbrace(q)_"Heat flow" + underbrace(w)_"work"

Thus,

$w = - q$

or for a reversible process (i.e. maximum efficiency),

${w}_{r e v} = - {\int}_{{V}_{1}}^{{V}_{2}} P \mathrm{dV} = - {q}_{r e v}$,

where the work is negatively-signed for work done by the system onto the surroundings.

This means that when the system of gas particles expands at constant temperature, the ability of the system to expand was due to the heat energy acquired, i.e. all the heat flowing in goes into pressure-volume work and does not change the temperature.

An example is if you have $\text{1 mol}$ of an ideal gas that reversibly expands to double its volume at $\text{298.15 K}$.

Then, the reversible work that gave rise to that expansion is found using the ideal gas law for the pressure:

${w}_{r e v} = - {\int}_{{V}_{1}}^{2 {V}_{1}} \frac{n R T}{V} \mathrm{dV}$

$= - n R T \ln \left(\frac{2 {V}_{1}}{V} _ 1\right) = - n R T \ln 2$

$= - \text{1.00 mols" xx "8.314472 J/mol"cdot"K" xx "298.15 K} \times \ln 2$

$= - \text{1718.28 J}$

So, the heat flowing in to perform that expansion would be

$\textcolor{b l u e}{{q}_{r e v}} = - {w}_{r e v} = \textcolor{b l u e}{+ \text{1718.28 J}}$