Under the condition of Constant temperature we say that the change in internal energy is 0.
Is the change in internal energy (at constant temperature) 0 for only ideal gases or substances in general ?
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1$\begingroup$ In general true only for ideal gases. Its due to the assumptions of ktg and dependence of velocity on temperature in ideal gases. $\endgroup$– Nisarg joshiCommented Dec 10, 2016 at 11:41
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1$\begingroup$ There are also other kinds of ideal substances for which this applies: incompressible solids and liquids $\endgroup$– Chet MillerCommented Dec 10, 2016 at 12:24
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1 Answer
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From the reciprocity relation for gases, the change in internal energy at constant temperature has the general relationship,
$$ \left(\frac{\partial e}{\partial v}\right)_T = -p + T \left(\frac{\partial p}{\partial T}\right)_v $$
where $e$ and $v$ is the internal energy and specific volume, respectively. Hence, only when $T \left(\frac{\partial p}{\partial T}\right)_v = p$, is the change in internal energy at constant temperature exactly zero. In the case of ideal gases, we have the thermal equation of state,
$$ p = \rho R T = \frac{R T}{v} $$
$$ \left(\frac{\partial p}{\partial T}\right)_v = \frac{R}{v} $$ which assumes the specific gas constant, $R$, is also independent of temperature. Substituting back in we have,
$$ \left(\frac{\partial e}{\partial v}\right)_T = -p + \frac{RT}{v} = -p + p $$ $$ \left(\frac{\partial e}{\partial v}\right)_T = 0 $$ Thus for gases, only when the ideal thermal equation of state is applicable does the change in internal energy at constant temperature exactly equal zero. In practice, this type of gas is referred to as a calorically perfect gas.
I don't usually work with liquids or solids, but I believe similar arguments can be made to justify under what conditions does the change in internal energy of a substance at constant temperature exactly equal zero.
