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 $T$ increases, as follows.
(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, $d V_{1}$ , in terms of $d T$ and the thermal expansion coefficient $β$ introduced in Problem 1.7.
(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, $d V_{2}$ , in terms of $d P$ and the isothermal compressibility $κ_{T}$ , defined as
$κ_{T} \equiv - \frac{1}{V} {(\frac{\partial V}{\partial P})}_{T}$
(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of $d P to d T$ is equal to $(\partial P / \partial T)_{V}$ , since there is no net change in volume. Express this partial derivative in terms of $β and κ_{T}$ . Then express it more abstractly in terms of the partial derivatives used to define $β and κ_{T}$ . For the second expression you should obtain
${(\frac{\partial P}{\partial T})}_{V} = - \frac{(\partial V / \partial T)_{P}}{(\partial V / \partial P)_{T}}$
This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.
(d) Compute $β, κ_{T}, and (\partial P / \partial T)_{V}$ for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).
(e) For water at $25^{\circ} C, β = 2.57 \times 10^{- 4} K^{- 1} and κ_{T} = 4.52 \times 10^{- 10} {Pa}^{- 1}$ . Suppose you increase the temperature of some water from $20^{\circ} C to 30^{\circ} C$ . How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which $(at 25^{\circ} C) β = 1.81 \times 10^{- 4} K^{- 1}$ and $κ_{T} = 4.04 \times 10^{- 11} {Pa}^{- 1}$
Given the choice, would you rather measure the heat capacities of these substances at constant $v$ or at constant $p$ ?

Question

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 $T$ increases, as follows.
(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume, $d V_{1}$ , in terms of $d T$ and the thermal expansion coefficient $β$ introduced in Problem 1.7.
(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, $d V_{2}$ , in terms of $d P$ and the isothermal compressibility $κ_{T}$ , defined as
$κ_{T} \equiv - \frac{1}{V} {(\frac{\partial V}{\partial P})}_{T}$
(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of $d P to d T$ is equal to $(\partial P / \partial T)_{V}$ , since there is no net change in volume. Express this partial derivative in terms of $β and κ_{T}$ . Then express it more abstractly in terms of the partial derivatives used to define $β and κ_{T}$ . For the second expression you should obtain
${(\frac{\partial P}{\partial T})}_{V} = - \frac{(\partial V / \partial T)_{P}}{(\partial V / \partial P)_{T}}$
This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.
(d) Compute $β, κ_{T}, and (\partial P / \partial T)_{V}$ for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).
(e) For water at $25^{\circ} C, β = 2.57 \times 10^{- 4} K^{- 1} and κ_{T} = 4.52 \times 10^{- 10} {Pa}^{- 1}$ . Suppose you increase the temperature of some water from $20^{\circ} C to 30^{\circ} C$ . How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which $(at 25^{\circ} C) β = 1.81 \times 10^{- 4} K^{- 1}$ and $κ_{T} = 4.04 \times 10^{- 11} {Pa}^{- 1}$
Given the choice, would you rather measure the heat capacities of these substances at constant $v$ or at constant $p$ ?

Accepted Answer

(A) The change in volume in $d V_{1}$ and thermal coefficient is $d V_{1} = β V d T$

(B) The change in volume of $d V_{2}$ is $d V_{2} = - κ_{T} V d P$

(C) The second expression is ${(\frac{\partial P}{\partial T})}_{V} = - \frac{(\partial V / \partial T)_{P}}{(\partial V / \partial P)_{T}}$

(D) An ideal gas of three expression is $β = \frac{1}{T}, κ_{T} = \frac{1}{P} . \frac{P}{T} = \frac{β}{κ_{T}}$

(E) The heat capacities of substances constant is $Δ P_{water} = 5.686 \times 10^{6} Pa, Δ P_{mercury} = 4.48 \times 10^{7} Pa$

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An Introduction to Thermal Physics

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Step :1 The thermal expansion coefficient (part a)

Step :2 Constant temperature (part b)

Step :3 Change in volume after two action (part c)

Step :4 Ideal gas law (part d)

Step :5 For water (part e)

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An Introduction to Thermal Physics

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