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This question already has an answer here:

For any adiabatic process $P$,$V$ and $T$ change while $dQ=0$. By first law of thermodynamics,

$$dQ=dU+dW$$

That is, it becomes

$$dU+dW=0$$

At this stage, we set

$$dU=C_v dT$$

Any further calculations are on this basis. If, in an adiabatic process, volume is not constant, then why do we write the above equation?

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marked as duplicate by stafusa, John Rennie thermodynamics Sep 20 '17 at 6:27

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    $\begingroup$ I think this equation is valid only for an ideal gas, but I am not sure. At least for an ideal gas one can proof that the internal energy is a function of only the temperature. Have a look in Reif's "Fundamentals of Statistical and Thermal Physics". $\endgroup$ – Semoi Sep 19 '17 at 18:17
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For an ideal gas, the internal energy is a function only of temperature. So it doesn't matter if the volume or pressure changes.

In freshman physics they taught us that $dQ=C_VdT$, but they didn't tell us that this equation only applies when no work is done. If work is done, then dQ is not equal to $C_vdT$. However, in thermodynamics, we got more precise (and smarter) by defining $C_v$ in terms of U (a state function that is independent of process path) rather than Q (a quantity that is intimately related to process path, and thus, not a physical property of the material): $$C_v=\left(\frac{\partial U}{\partial T}\right)_V\tag{1}$$This relationship is consistent with the freshman physics definition if the system volume is constant, such that $Q=\Delta U=C_v\Delta T$. But, Eqn. 1 is much more general than that and helps us determine the change in internal energy between any arbitrary pair of thermodynamic equilibrium states. For example, for an ideal gas, if we wish to determine the change in internal energy U between two states with different temperatures and volume, we can first integrate Eqn. 1 at constant volume between the initial and final temperatures and then determine the change at constant final temperature if we go from the initial to the final volume. But, the latter is zero, since U doesn't depend on volume for an ideal gas.

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This is a well-known source of confusion. Try looking at it like this… For an ideal gas the internal energy depends on the temperature alone, and the relationship is one of proportionality, so $$dU=constant \times dT.$$ For example, for an ideal monatomic gas, $constant=\frac{3}{2}nR$, in which n is the number of moles. The constant is therefore a constant for the sample of gas, independent of any process the gas is going through. The heat capacity at constant volume, $n c_v$, works out to be equal to that constant, because, at constant volume, $W=0$, so $Q=dU$, so $n c_v dT = constant \times dT$. Therefore $n c_v=constant$, so we can call the constant "$n c_v$" even though the constant is independent of any particular process the gas is going through!

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First we need to understand why $dU = m c_v dT$.

The definition of specific heat comes from the following formulation: $$c_{y} = \frac{1}{m} \left(\frac{\delta Q}{d T} \right)_{y} $$ note that specific heat depends on the process $y$.

Let's now consider the first law of thermodynamics: $$ \delta Q + \delta W = d U \\ \delta Q = d U - \delta W\\ \delta Q = d U - p d V $$

Let's apply this for a constant volume ($dV=0$) process: $$ (\delta Q)_v = d U $$

dividing both sides by $dT$: $$\left(\frac{\delta Q}{d T} \right)_v = \frac{d U}{d T}$$

the index $v$ is used because the process is for a constant volume.

Now, using the definition of specific heat, we arrive at: $$ c_v = \frac{1}{m} \left(\frac{\delta Q}{d T} \right)_v \\ c_v = \frac{1}{m} \frac{d U}{d T} $$

One should note that the specific heat at constant volume only depends on properties $U$ and $T$. Since it only depends on state properties, it means that it does not depend on the process. For this particular reason it can be applied to any process.

A similar procedure can be done to explain that the specific heat at constant pressure can also be applied to any process. $$ c_p = \frac{1}{m} \frac{d H}{d T} $$

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  • $\begingroup$ For a real gas, both Cv and U also depend on volume. $\endgroup$ – Chet Miller Sep 19 '17 at 19:04

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