Q.5.14 The partial-derivative relations... [FREE SOLUTION]

Chapter 5: Q.5.14 (page 159)

The partial-derivative relations derived in Problems 1.46,3.33, and 5.12, plus a bit more partial-derivative trickery, can be used to derive a completely general relation between $C_{P}$ and $C_{V}$ .
(a) With the heat capacity expressions from Problem 3.33 in mind, first consider $S$ to be a function of $T$ and $V .$ Expand $d S$ in terms of the partial derivatives $(鈭� S / 鈭� T)_{V}$ and $(鈭� S / 鈭� V)_{T}$ . Note that one of these derivatives is related to $C_{V}$
(b) To bring in $C_{P}$ , considerlocalid="1648430264419" $V$ to be a function of $T$ and P and expand dV in terms of partial derivatives in a similar way. Plug this expression for dV into the result of part (a), then set $d P = 0$ and note that you have derived a nontrivial expression for $(鈭� S / 鈭� T)_{P}$ . This derivative is related to $C_{P}$ , so you now have a formula for the difference $C_{P} - C_{V}$
(c) Write the remaining partial derivatives in terms of measurable quantities using a Maxwell relation and the result of Problem 1.46. Your final result should be
$C_{P} = C_{V} + \frac{T V 尾^{2}}{魏_{T}}$
(d) Check that this formula gives the correct value of $C_{P} - C_{V}$ for an ideal gas.
(e) Use this formula to argue that $C_{P}$ cannot be less than $C_{V}$ .
(f) Use the data in Problem 1.46 to evaluate $C_{P} - C_{V}$ for water and for mercury at room temperature. By what percentage do the two heat capacities differ?
(g) Figure 1.14 shows measured values of $C_{P}$ for three elemental solids, compared to predicted values of $C_{V}$ . It turns out that a graph of $尾$ vs.T for a solid has same general appearance as a graph of heat capacity. Use this fact to explain why $C_{P}$ and $C_{V}$ agree at low temperatures but diverge in the way they do at higher temperatures.

Short Answer

Expert verified

a) $d S = {(\frac{鈭� S}{鈭� V})}_{T} d V + {(\frac{鈭� S}{鈭� T})}_{V} d T$

b) $C_{P} = T {(\frac{鈭� S}{鈭� V})}_{T} {(\frac{鈭� V}{鈭� T})}_{P} + C_{V}$

c) $C_{P} = \frac{尾^{2} T V}{魏_{T}} + C_{V}$

d) $C_{P} = C_{V} + N k$

e) $C_{P} > C_{V}$

f) $C_{P} - C_{V} = 0.0435 J / K, C_{P} - C_{V} = 3.576 J / K$

g) $C_{P} - C_{V} 鈭� T 尾^{2}$

Step by step solution

Part (a) - Step 1: Solution

Let S be a function of V and T, that is $S = S (V, T)$ Then, based on the derivative's definition, we get

role="math" localid="1648427685193" $d S = {(\frac{鈭� S}{鈭� V})}_{T} d V + {(\frac{鈭� S}{鈭� T})}_{V} d T . . . . . . (1)$

the heat capacity at constant volume and at constant pressure are given by:

role="math" localid="1648427702735" $C_{V} = T {(\frac{鈭� S}{鈭� T})}_{V} C_{P} = T {(\frac{鈭� S}{鈭� T})}_{P} . . . . . . (2)$

so the first term of equation (1), is $C_{V} / T$ .

Part(b) Step 2: Solution

Let V be a function of P and T, that is $V = V (P, T)$ , then from the definition of the derivative we get:

$d V = {(\frac{鈭� V}{鈭� P})}_{T} d P + {(\frac{鈭� V}{鈭� T})}_{P} d T$

at constant pressure $d p$ the above equation becomes

$d V = {(\frac{鈭� V}{鈭� T})}_{P} d T$

Substitute in equation (1) to get:

$(d S)_{P} = {(\frac{鈭� S}{鈭� V})}_{T} {(\frac{鈭� V}{鈭� T})}_{P} d T + {(\frac{鈭� S}{鈭� T})}_{V} d T {(\frac{鈭� S}{鈭� T})}_{P} = {(\frac{鈭� S}{鈭� V})}_{T} {(\frac{鈭� V}{鈭� T})}_{P} + {(\frac{鈭� S}{鈭� T})}_{V}$

Multiply T on the both side we get:

$T {(\frac{鈭� S}{鈭� T})}_{P} = T {(\frac{鈭� S}{鈭� V})}_{T} {(\frac{鈭� V}{鈭� T})}_{P} + T {(\frac{鈭� S}{鈭� T})}_{V}$

Substitute from equation (2) we get:

role="math" localid="1648428387185" $C_{P} = T {(\frac{鈭� S}{鈭� V})}_{T} {(\frac{鈭� V}{鈭� T})}_{P} + C_{V} . . . . . . (3)$

Part (c) - Step 3: Solution

Consider the Helmholtz energy Maxwell relation we have:

${(\frac{鈭� S}{鈭� V})}_{T} = {(\frac{鈭� P}{鈭� T})}_{V}$

substitute into (3) we get:

$C_{P} = T {(\frac{鈭� P}{鈭� T})}_{V} {(\frac{鈭� V}{鈭� T})}_{P} + C_{V}$

The following is the relationship between thermal expansion and isothermal compressibility:

$\frac{尾}{魏_{T}} = {(\frac{鈭� P}{鈭� T})}_{V}$

Substitute in above expression:

$C_{P} = \frac{尾 T}{魏_{T}} {(\frac{鈭� V}{鈭� T})}_{P} + C_{V}$

The thermal expansion is determined by:

$尾 = \frac{1}{V} {(\frac{鈭� V}{鈭� T})}_{P}$

Substitute value in above equation we get:

$C_{P} = \frac{尾^{2} T V}{魏_{T}} + C_{V} . . . . . . (4)$

Part (d) - Step 4: Solution

$P V = N k T$ is the ideal gas relation, hence the thermal expansion and isothermal compressibility are given by:

$尾 = \frac{1}{V} {(\frac{鈭� V}{鈭� T})}_{P} = \frac{1}{V} \frac{鈭�}{鈭� T} {(\frac{N k T}{P})}_{P} = \frac{N k}{P V} = \frac{1}{T} 魏_{T} = - \frac{1}{V} {(\frac{鈭� V}{鈭� P})}_{T} = \frac{1}{V} \frac{鈭�}{鈭� P} {(\frac{N k T}{P})}_{T} = \frac{N k T}{P^{2} V}$

substitute into (4) to get:

$C_{P} = \frac{T V}{T^{2}} \frac{P^{2} V}{N k T} + C_{V} C_{P} = \frac{P^{2} V^{2}}{N k T^{2}} + C_{V}$

To use $P V = N K T$ we get:

$C_{P} = \frac{N^{2} k^{2} T^{2}}{N k T^{2}} + C_{V} C_{P} = C_{V} + N k$

Part (e) - Step 5: Solution

The thermal expansion can be negative, but it is squared, it will always be positive; however, the thermal compressibility cannot be negative because adding pressure to the system will increase its volume, which is impossible; hence, $C_{P}$ is always greater than $C_{V}$ .

Part (f) - Step 6: Solution

Using the data from problem 1.46e The difference between the heat capacities at constant pressure and constant volume for one gram of water at room temperature is:

$C_{P} - C_{V} = \frac{尾^{2} T V}{魏_{T}} = \frac{{(2.57 脳 10^{- 4} K^{- 1})}^{2} (298 K) (10^{- 6} m^{3})}{4.52 脳 10^{- 10} {Pa}^{- 1}} = 0.0435 J / K$

role="math" localid="1648429880211" $this is just 1 % of the heat capacity 4.186 J / K . Now for one mole of mercury (from the data at page 405) we have:$

$C_{P} - C_{V} = \frac{尾^{2} T V}{魏_{T}} = \frac{{(1.81 脳 10^{- 4} K^{- 1})}^{2} (298 K) (14.8 脳 10^{- 6} m^{3})}{4.04 脳 10^{- 11} {Pa}^{- 1}} = 3.576 J / K$

$this is just under 13 % of the heat capacity C_{P} = 28.0 J / K$

Part (g) - Step 7: Solution

Since the volume of a solid is not totally dependent by its temperature,

$C_{P} - C_{V} 鈭� T 尾^{2}$

As a result, the difference in heat capacity as the temperature approaches zero is zero, and the two curves of $C_{P}$ and $C_{V}$ diverge at high temperatures.

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Short Answer

Step by step solution

Part (a) - Step 1: Solution

Part(b) Step 2: Solution

Part (c) - Step 3: Solution

Part (d) - Step 4: Solution

Part (e) - Step 5: Solution

Part (f) - Step 6: Solution

Part (g) - Step 7: Solution

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