Entropy of the universe

The entropy of the universe tends to increase with time. To understand this statement, we need to study the change in entropy of adiabatic processes.

For a reversible adiabatic process in an isolated system, $dq_{rev}=0$ and from eq119, $dS=0$ . Therefore,

$\int_cdS=\Delta S=0\;\;\;\;\;\;\;\;(128)$

For an irreversible adiabatic process AB with well-defined start and end states (see diagram above), we can set up two reversible paths, BC and CA, to create a cyclic process ABC where BC is a reversible isochoric process and CA is a reversible adiabatic process. The change in entropy for the cycle is:

$\oint dS_{sys}=\int_A^BdS_{AB}+\int_B^CdS_{BC}+\int_C^AdS_{CA}$

Since entropy is a state function, $\oint dS_{sys}=0$ and because CA is a reversible adiabatic process, $dS_{CA}=0$ . So,

$\int_A^BdS_{AB}=-\int_B^CdS_{BC}$

Since BC is a reversible process, we can substitute eq119 in $dS_{BC}$ to give

$\int_A^BdS_{AB}=-\int_{T_B}^{T_C}\frac{dq}{T}$

From eq42, $dq=C_VdT$ since BC is an isochoric process,

$\int_A^BdS_{AB}=-\int_{T_B}^{T_C}\frac{C_V}{T}dT$

As $T_C<T_B$ and $C_V>0$ , $\int_A^BdS_{AB}>0$ and therefore

$\Delta S>0\;\;\;\;\;any\;irreversible\;adiabatic\;process,\; isolated\;system\;\;\;\;\;\;(129)$

Combining eq128 and eq129

$\Delta S\geq 0\;\;\;\;\;any\;adiabatic\;process,\; isolated\;system\;\;\;\;\;\;(130)$

If we regard our physical universe as an isolated system with any process occurring within it as adiabatic,

$\Delta S_{uni}> 0\;\;\;\;\;any\;irreversible\;process\;\;\;\;\;\;(131)$

$\Delta S_{uni}= 0\;\;\;\;\;any\;reversible\;process\;\;\;\;\;\;(132)$

$\Delta S_{uni}\geq 0\;\;\;\;\;any\;process\;\;\;\;\;\;(133)$

Since an irreversible process is a spontaneous process,

$\Delta S_{uni}> 0\;\;\;\;\;any\;spontaneous\;process\;\;\;\;\;\;(134)$

We can therefore also define reversibility and irreversibility (or spontaneity) in terms of entropy of the universe, where a process is reversible if there is no change in entropy of the universe, and where a process is irreversible (or spontaneous) if the change in entropy of the universe is positive. As mentioned in the article on reversible and irreversible processes, a reversible process is an idealisation, which makes all real processes irreversible (or spontaneous). Hence, the consequence of eq131 is that the entropy of the universe tends to increase with time. If there comes a time when no more irreversible process in the universe occurs, equilibrium is attained and $dS_{uni}=0$ . This hypothesis is known as the heat death of the universe.

Question

Show that $dq=C_VdT$ for an isochoric process.

Answer

The change in internal energy for any process that involves only pV work is given by

$dU=dq-pdV\;\;or\;\;dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$

For an isochoric process, $dV=0$ , so $dU=dq$ or $dU=\left(\frac{\partial U}{\partial T}\right)_VdT$ . Hence, $dq=\left(\frac{\partial U}{\partial T}\right)_VdT=C_VdT$ .