PX154 - E6b - quantitative definition of entropy

a small change in entropy, $d S$ , is given by a small amount of heat transfer, $d Q$ , at a temperature, $T$ :

d S = \frac{d Q}{T} $ $ [Y F e q n 20.1] - i f a s m a l l a m o u n t o f h e a t, $ d Q $, i s p u t i n t o a s y s t e m : $ $ d Q = d U + d W

- $d U \to$ increase in internal energy $⟹$ increases randomness, and hence entropy.
- $d W = p d V \to$ increase volume, molecules have more space to move in, and hence entropy increases

key notes:
- the relationship given only holds for reversible processes
- for an isothermal process, $T$ constant, we can write $Δ S = \frac{Q}{T}$
$\begin{align*} Q = W &= nRT \ln{\frac{V_{2}}{V_{1}}} \\ \Delta S &= nR\ln{\frac{V_{2}}{V_{1}}}$
- for an irreversible process, $Δ S$ can still be found using a reversible process that has the same initial and final states.
eg: heat $1 k g$ water from $0 ° C \to 100 ° C$ . what is the change in entropy?
- heat capacity of water, $c_{m a s s} = 4190 J K^{- 1} k g^{- 1}$
- $T$ is not constant
$Δ S = \int_{s t a t e 1}^{s t a t e 2} \frac{d Q}{T}$
- heat, $Q$ , is supplied to heat the water
- $c_{m a s s} = \frac{1}{m} \frac{d Q}{d T}$
- so, $d Q = m c_{m a s s} d T$
$\begin{align*} \Delta S &= \int_{T_{1}}^{T_{2}} \frac{m c_{mass}}{T}\,dT \\ &= m c_{mass}\ln\frac{T_{2}}{T_{1}}$ $∴ Δ S = 1.31 \times 10^{3} J K^{- 1}$
cooling back down to $0 ° C$ , $Δ S = - 1.3 \times 10^{3} J K^{- 1}$
- ie: $Δ S = 0$ for $0 ° C \to 100 ° C \to 0 ° C$
this is the entropy change of just the water
what about the entropy change of the universe?
- if a reversible process is used to heat and cool the water, $0 ° C \to 100 ° C$ , then, $Δ S_{u n i v e r s e} = 0$
- however, if the water is heated on a bunsen burner, and cooled in a fridge the process will be irreversible:

\begin{aligned} Δ S_{w a t e r} & = 0 \\ Δ S_{u n i v e r s e} & > 0 \end{aligned}

for all real processes:

Δ S_{u n i v e r s e} > 0

furthermore, for an isolated system:

Δ S \geq 0

comment:
- in the carnot cycle, the cooling must be adiabatic
  - what about isochoric cooling?
    - if only the gas is considered, it can be cooled and heated between $A$ and $B$ reversibly
    - but in the engine, the "isolated system" must include the reservoir
- what about the example in the last section?
  - irreversible (will not spontaneously retain)
  - can use two reversible processes:
    - cool hot object from $100 ° C \to 50 ° C$
    - heat cold object to $50 ° C$
  - calculate $Q$ for each process (heat loss = heat gain)
  - calculate $Δ S$ for each region and add to get the total