PC2352 Examples 4

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7.

By rearranging the fundamental thermodynamic relation to give ${\rm d}S$ in terms of ${\rm d}E$ and ${\rm d}V$, show that for an ideal gas the change in entropy is given by

$$\Delta S=n R\ln\left({V_2\over V_1}\right)+n c_{\scriptscriptstyle V}\ln\left({T_2\over T_1}\right),$$ (5.1)

where $c_{\scriptscriptstyle V}$ is the molar heat capacity at constant volume.

A kilo of steam starts at a temperature of 250$^{\rm o}$C and 1.4 atmospheres, and expands and cools to 150$^{\rm o}$C and 1 atmosphere. The surroundings are at 20$^{\rm o}$C and 1 atmosphere. What is the maximum amount of useful work that can be extracted during the expansion? (Treat the steam as an ideal gas, and take the molar heat capacity $c_{\scriptscriptstyle V}$ to be 27Jmol$^{-1}$K$^{-1}$.)

8.

At 25$^{\circ}$C the volume of water, $V$, is well approximated by the formula

$$V = \Big(1.80689 \times 10^{-5} - 8.06 \times 10^{-10} P + 1.3 \times 10^{-14} P^2\Big) \hbox{m$^3$ mol$^{-1}$}$$

for $P$ between 1 and 1000 atmospheres and

$$\left( \frac{\partial V}{\partial T} \right)_P =\Big( 4.69 \... ... \times 10^{-12} P \Big) \hbox{m$^3$ K$^{-1}$ mol$^{-1}$}$$

In both cases $P$ is measured in atmospheres.
The data comes from the water property calculator at the University of Washington Earth Science website. The first term in $V$ is the molar volume in m$^3$ and the ratio of the first to the second term is the isothermal bulk modulus ( $=V (\partial P/\partial V)_{\scriptscriptstyle T}$) in atmospheres.

Using the first equation, find the work necessary to perform a reversible isothermal compression of 1 mole of water from 1 atmosphere to 1000 atmospheres at 25$^{\circ}$C. Using the second equation and a Maxwell relation, calculate the decrease in the internal energy during this process.
Hint: the Maxwell relation allows us to find the entropy change $\Delta S$, and as the process is isothermal we can therefore find $Q$. Watch that final results are in "atm m$^{3}$" which needs multiplied by $1.01325\times 10^5$ to convert to Joules.
Compare with qu. (1.iii) where we couldn't find the heat given out because we didn't have the data above.

9.

For one mole of a gas obeying the van der Waals' equation,

$$P = \frac{RT}{V-b} - \frac{a}{V^{2}} :$$

(i)

Using the fundamental thermodynamic relation and a suitable Maxwell relation, show that

$$\left( \frac{\partial E}{\partial V} \right)_T = \frac{a}{V^{2}}$$

(ii)

Hence show that $C_{\scriptscriptstyle V}$ depends only on $T$, i.e. $(\partial C_{\scriptscriptstyle V}/\partial V)_{\scriptscriptstyle T}= 0$
Hint: show first that, since at constant volume ${}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}W^{\rm rev}=0$, $C_{\scriptscriptstyle V}=(\partial E/\partial T)_{\scriptscriptstyle V}$.

(iii)

By writing $S$ as a function of $T$ and $V$, show that

$${\rm d}S = \frac{R {\rm d}V}{V-b} + \frac{C_{\scriptscriptstyle V}{\rm d}T}{T}$$

and hence find the analogue of the ideal gas relation $TV^{\gamma-1}$= const. for a reversible adiabatic process involving a van der Waal gas. (Take $C_V$ to be constant.)
See here

(iv)

Obtain the isobaric coefficient of thermal expansion $\alpha$

$$\alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P$$

and hence, using expressions from lecture notes, calculate $C_{\scriptscriptstyle P}- C_{\scriptscriptstyle V}$.
Hint: for $\alpha$ use implicit differentiation rather than solving for $V$ first, and note that $(\partial P/\partial T)_{\scriptscriptstyle P}=0$. The answer IS messy! Check your answer for $C_{\scriptscriptstyle P}- C_{\scriptscriptstyle V}$ here.