Heat and entropy

Heat and Entropy

At constant volume ( $d W = 0$ ), $d Q = d E$ . Using the definition of temperature:

d Q = d E = {(\frac{\partial \log Ω}{\partial E})}^{- 1} d (\log Ω) = T d S

So adding heat raises entropy; removing heat lowers it.

For heat $d Q$ flowing from A to B:

d S = d S_{B} + d S_{A} = \frac{d Q}{T_{B}} - \frac{d Q}{T_{A}}

$T_{A} = T_{B}$ : $d S = 0$ , no heat flows (thermal equilibrium)
$T_{A} < T_{B}$ : $d S < 0$ , forbidden by the second law
$T_{A} > T_{B}$ : $d S > 0$ ✓ — heat flows from hot to cold, total entropy increases

Proportionality between heat added and temperature rise: $C \equiv \frac{d Q}{d T}$

At constant volume (no work done):

C_{V} = T {(\frac{\partial S}{\partial T})}_{V}

At constant pressure (system can expand, $d E = T d S - P d V$ ):

C_{P} = T {(\frac{\partial S}{\partial T})}_{P} + P {(\frac{\partial V}{\partial T})}_{P}

$C_{P} > C_{V}$ for nearly all systems — at constant pressure, some heat goes into expansion work rather than raising temperature.

Ideal gas: non-interacting particles, $Ω \propto V^{N}$ . From the definition of pressure:

P = k T \frac{\partial \log Ω}{\partial V} = N k T \frac{\partial \log V}{\partial V} = \frac{N k T}{V}

P V = N k T

Fix $V$ : pressure scales with $T$
Fix $P$ : gas expands with $T$ ; expansion does work $W = P Δ V$ , so temperature tends to drop
Good approximation for dilute, non-polar, non-reactive gases (e.g. air)