# Isothermal change of state

Comparison of the isothermal and isentropic changes in state in the pV diagram

The isothermal change of state is a thermodynamic change of state in which the temperature remains unchanged:

${\ displaystyle T = {\ text {const.}} \ quad \ Leftrightarrow \ quad T_ {1} = T_ {2}}$

There denote and the temperatures before and after the change of state. When a gas is compressed, the heat of compression has to be dissipated, and when it is expanded, heat has to be added ( diabatic change of state ) . This can be achieved approximately by a heat bath . ${\ displaystyle T_ {1}}$${\ displaystyle T_ {2}}$

## Ideal gas

According to Boyle-Mariotte's law and the caloric equation of state of an ideal gas , the product of pressure and volume as well as the internal energy remain constant at constant temperature : ${\ displaystyle T}$ ${\ displaystyle p}$ ${\ displaystyle V}$ ${\ displaystyle U}$

${\ displaystyle p \ cdot V = n \ cdot R \ cdot T = {\ text {const.}} \ quad \ Leftrightarrow \ quad p \ sim {\ frac {1} {V}}}$.

From this it follows that the pressures are inversely proportional to the corresponding volumes:

${\ displaystyle {\ frac {V_ {1}} {V_ {2}}} = {\ frac {p_ {2}} {p_ {1}}}}$

For the work done, the following applies to isothermal compression or expansion of moles of an ideal gas: ${\ displaystyle \ W}$${\ displaystyle n}$

${\ displaystyle \ W = n \, R \, T_ {1} \ ln \ left ({\ frac {V_ {1}} {V_ {2}}} \ right) = n \, R \, T_ {1 } \ ln \ left ({\ frac {p_ {2}} {p_ {1}}} \ right) = p_ {1} \, V_ {1} \ ln \ left ({\ frac {V_ {1}} {V_ {2}}} \ right)}$,

where denotes the universal gas constant . ${\ displaystyle R}$

Because is . According to the first law of thermodynamics ( ) it follows that the added or extracted heat corresponds directly to the work performed ( ). ${\ displaystyle T_ {2} = T_ {1}}$${\ displaystyle \ Delta U = 0}$${\ displaystyle \ Delta U = Q + W}$${\ displaystyle \ Q = -W}$