I don't understand why the change in internal energy in an isothermal expansion is considered to be 0. In an isothermal process is done work, so, shouldn't that imply that the internal energy changed? Isn't it necessary that the work done should result in a change of internal energy? Or, does it mean that internal energy only changes if there is a gain in heat $dQ$?

  • $\begingroup$ "change in internal energy in an isothermal expansion considered to be 0" only for an ideal gas, and was experimentally verified by Joule for low density real gases at high enough temperature. $\endgroup$ – hyportnex Feb 6 '15 at 14:30
  • $\begingroup$ For an ideal gas, all the internal energy is kinetic (well, if there's no interaction between particles, there is no potential energy). Kinetic energy only depends on temperature, thus it's energy is conserved during the process. This indeed implies (I think you theorized about something similar in your question), that exactly the amount of work done on the gas will be lost for the gas due to heat exchange (or vice versa). $\endgroup$ – kristjan Feb 6 '15 at 21:35

in an isothermal process work is done ryt..then so if work is done dosent it mean that there is a change In internal energy

Remember what internal energy $U$ is. It is the sum:

  • of atomic vibrations (sensed as the temperature $T$),
  • of chemical bonds or "binding energy" (this includes phase changes from e.g. liquid to solid, where internal energy is lowered but temperature is constant),
  • of potential energy (if the object is at a high shelf, it has a "potential" to do work),
  • of kinetic energy (only if it moves $K=½mv^2$),
  • and others alike.

But, doing work $W$ on an object could e.g. be displacing it sideways a distance $x$ - that is, putting it somewhere else. This would require some force $F$, and:

$$W=F \cdot x$$

In this new position no changes are done in chemical composition, no changes in potential energy, no kinetic energy (it lies still on the table), and no temperature change (since you want it to be a isothermal process). So this is an example of work done with no changes in internal energy.

For an isothermal process temperature has to be constant. After you have displaced (pushed) the box and done work on it, friction might stop it. This would generate heat. For an isothermal process this heat $Q$ has to be removen right away (maybe you use a cooling system or liquid). Now, conservation of energy from thermodynamics means that:

$$\Delta U = Q-W$$

And since - as discussed above - internal energy is unchanged, $\Delta U=0$ and $W=Q$.

  • 1
    $\begingroup$ Another explanation could be given by deriving the formula for change in internal energy to be given by $dU=\frac{f}{2}nRdT$ and then in an isothermal process $dT=0$ and hence the result. $\endgroup$ – ritvik1512 Feb 6 '15 at 15:25

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