solvwe allProblem 1.46. Measured heat capacities of solids and liquids are almost always
at constant pressure, not constant volume. To see why, estimate the pressure
needed to keep V fixed as 7" inc

Question

solvwe allProblem 1.46. Measured heat capacities of solids and liquids are almost always
at constant pressure, not constant volume. To see why, estimate the pressure
needed to keep V fixed as 7" increases, as follows.

(a) First imagine slightly increasing the temperature of a material at constant
pressure. Write the change in volume, dV, in terms of d7" and the thermal
expansion coefficient 3 introduced in Problem 1.7.

(b) Now imagine slightly compressing the material, holding its temperature
fixed. Write the change in volume for this process, dV3, in terms of dP and
the isothermal compressibility x7, defined as

—_—
RAC
(This is the reciprocal of the isothermal bulk modulus defined in Prob-
lem 1.39.)

(c) Finally, imagine that you compress the material just enough in part (b)
to offset the expansion in part (a). Then the ratio of dP to dT is equal
to (OP/8T)y, since there is no net change in volume. Express this partial
derivative in terms of 3 and x7. Then express it more abstractly in terms of
the partial derivatives used to define 3 and xp. For the second expression
you should obtain

aP\ _ _(av/aT)p
(3), ~ vier)’
This result is actually a purely ma

VIPSolutions · Accepted Answer

Let’s solve each part step by step:

---

### (a) **Change in Volume at Constant Pressure**

At constant pressure, if you increase the temperature by $ dT $, the change in volume is given by the **thermal expansion coefficient** $ \beta $:

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

So, the change in volume:

$$
dV_1 = \left(\frac{\partial V}{\partial T}\right)_P dT = \beta V dT
$$

---

### (b) **Change in Volume at Constant Temperature**

At constant temperature, if you change the pressure by $ dP $, the change in volume is given by the **isothermal compressibility** $ \kappa_T $:

$$
\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T
$$

So, the change in volume:

$$
dV_2 = \left(\frac{\partial V}{\partial P}\right)_T dP = -\kappa_T V dP
$$

---

### (c) **Finding $\left(\frac{\partial P}{\partial T}\right)_V$ in Terms of $\beta$ and $\kappa_T$**

To keep volume constant, the total change in volume must be zero:

$$
dV_1 + dV_2 = 0
$$
$$
\beta V dT - \kappa_T V dP = 0
$$
$$
\beta dT = \kappa_T dP
$$
$$
\frac{dP}{dT} = \frac{\beta}{\kappa_T}
$$

So,

$$
\left(\frac{\partial P}{\partial T}\right)_V = \frac{\beta}{\kappa_T}
$$

For the abstract expression involving partial derivatives (from the problem):

$$
\left(\frac{\partial P}{\partial T}\right)_V = -\frac{(\partial V / \partial T)_P}{(\partial V / \partial P)_T}
$$

Recall:
- $ (\partial V / \partial T)_P = \beta V $
- $ (\partial V / \partial P)_T = -\kappa_T V $

So,

$$
-\frac{\beta V}{-\kappa_T V} = \frac{\beta}{\kappa_T}
$$

---

### (d) **For an Ideal Gas**

For an ideal gas:
- $ V = \frac{nRT}{P} $

Find each term:

#### $ \beta $:

$$
\beta = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P
= \frac{1}{V} \cdot \frac{nR}{P} = \frac{1}{T}
$$

#### $ \kappa_T $:

$$
\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = -\frac{1}{V} \cdot \left(-\frac{nRT}{P^2}\right) = \frac{V}{P} = \frac{1}{P}
$$

#### $ \left(\frac{\partial P}{\partial T}\right)_V $:

From the ideal gas law, $ P = \frac{nRT}{V} $, so:

$$
\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V} = \frac{P}{T}
$$

Check that:

$$
\frac{\beta}{\kappa_T} = \frac{1/T}{1/P} = \frac{P}{T}
$$

Which matches.

---

### (e) **Numerical Calculation for Water and Mercury**

Given:
- For water at 25°C:
  - $ \beta = 2.57 \times 10^{-4} $ K$^{-1}$
  - $ \kappa_T = 4.52 \times 10^{-10} $ Pa$^{-1}$

Temperature increase: $ \Delta T = 10 $ K.

The pressure required to offset the expansion:

$$
\frac{dP}{dT} = \frac{\beta}{\kappa_T}
$$
$$
\Delta P = \frac{\beta}{\kappa_T} \Delta T
$$

Plug in the values:

$$
\Delta P = \frac{2.57 \times 10^{-4}}{4.52 \times 10^{-10}} \times 10
= (5.685 \times 10^{5}) \times 10
= 5.69 \times 10^{6} \text{ Pa}
$$

#### For Mercury:
- $ \beta = 1.81 \times 10^{-4} $ K$^{-1}$
- $ \kappa_T = 4.04 \times 10^{-11} $ Pa$^{-1}$

$$
\Delta P = \frac{1.81 \times 10^{-4}}{4.04 \times 10^{-11}} \times 10
= (4.48 \times 10^{6}) \times 10
= 4.48 \times 10^{7} \text{ Pa}
$$

---

### (f) **Would You Rather Measure Heat Capacity at Constant V or P?**

It is much easier to keep pressure constant than to keep volume constant because the pressure required to keep the volume fixed changes drastically with small temperature changes (as calculated above). Thus, it is much more practical to measure heat capacities at constant pressure.

---

## **Summary Table**

| Substance | $\beta$ (K$^{-1}$) | $\kappa_T$ (Pa$^{-1}$) | $\Delta P$ for $ \Delta T = 10\,K $ (Pa) |
|-----------|------------------------|----------------------------|----------------------------------------------|
| Water     | $2.57 \times 10^{-4}$     | $4.52 \times 10^{-10}$         | $5.69 \times 10^6$                      |
| Mercury   | $1.81 \times 10^{-4}$     | $4.04 \times 10^{-11}$         | $4.48 \times 10^7$                      |

---

Let me know if you want a more detailed algebraic step for any part!

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