One mole of a gas obeys the van der Waals equation of state: $$ \left(P+\frac{a}{v^{2}}\right)(v-b)=R T $$ and its molar internal energy is given by $$ u=c T-\frac{a}{v}, $$ where \(a, b, c\), and \(R\) are constants. Calculate the molar heat capacities \(c_{V}\) and \(c_{P}\).

Short Answer

Expert verified
c_{V} = c, c_{P} = c + R.

Step by step solution

01

Understand Molar Heat Capacities

Molar heat capacity at constant volume, \(c_{V}\), is defined as \(c_{V} = \left( \frac{\partial u}{\partial T} \right)_{V}\). Molar heat capacity at constant pressure, \(c_{P} is defined as c_{P} = \left( \frac{\partial H}{\partial T} \right)_{P}\), where \(H\) is the enthalpy given by \(H = u + Pv\).
02

Calculate \(c_{V}\)

Given \(u = c T - \frac{a}{v}\), the partial derivative with respect to \(T\) at constant volume is: \(c_{V} = \left( \frac{\partial }{\partial T} (c T - \frac{a}{v}) \right)_{V} = c\).
03

Calculate Enthalpy \(H\)

Given \(H = u + Pv\), and using the van der Waals equation \(\left(P + \frac{a}{v^{2}}\right)(v - b) = R T\), the pressure can be written as \(P = \frac{R T}{v - b} - \frac{a}{v^{2}}\). Thus, \(H = u + \left(\frac{R T}{v - b} - \frac{a}{v^{2}}\right) v\).
04

Simplify Enthalpy Expression

Substitute the expression of \(u\) into the enthalpy expression to get: \(H = (c T - \frac{a}{v}) + \left(\frac{R T}{v - b} - \frac{a}{v^{2}}\right) v\). Simplify it to: \(H = c T + \frac{R T v}{v - b} - \frac{a b}{v (v - b)}\). For an ideal scenario (large \(v\)), \(H ≈ c T + R T\).
05

Calculate \(c_{P}\)

For the simplified case: \(c_{P} = \left( \frac{\partial H}{\partial T} \right)_{P} = c + R\). Therefore, \(c_{P} = c + R\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Van der Waals Equation
The van der Waals equation is an adjustment to the ideal gas law, accounting for the volume occupied by gas molecules and the attractive forces between them. The equation is given by:

\[ \bigg(P + \frac{a}{v^2}\bigg)(v - b) = RT \]

Here, \( P \) is the pressure, \( v \) is the molar volume, \( R \) is the gas constant, and \( T \) is the temperature. The constants \( a \) and \( b \) account for the intermolecular forces and the volume occupied by gas molecules, respectively.

This equation is essential for real gases, especially under high pressure and low temperature conditions where deviations from ideal behavior are significant. It helps to explain why real gases don't follow the ideal gas law perfectly.
Molar Internal Energy
Molar internal energy refers to the total energy contained within one mole of a substance. It includes kinetic and potential energy due to molecular motion and intermolecular forces. For a gas obeying the van der Waals equation:

\[ u = cT - \frac{a}{v} \]

Here, \( u \) is the molar internal energy, \( c \) is a specific heat constant, \( T \) is the temperature, and \( \frac{a}{v} \) accounts for intermolecular forces. The term \( cT \) represents the kinetic energy, while \( -\frac{a}{v} \) accounts for potential energy due to attractions between molecules.

In essence, molar internal energy helps us understand how temperature and volume changes affect the energy within a gas.
Enthalpy
Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is an essential concept in understanding heat transfer processes at constant pressure. Enthalpy \( H \) is given by:

\[ H = u + Pv \]

Using the van der Waals equation, we can replace \( P \) with:

\[ P = \frac{RT}{v - b} - \frac{a}{v^2} \]

Substituting this into the enthalpy equation:

\[ H = cT - \frac{a}{v} + \bigg(\frac{RT}{v - b} - \frac{a}{v^2}\bigg)v \]

Simplifying, we get:

\[ H = cT + \frac{RTv}{v - b} - \frac{ab}{v(v - b)} \]

For an ideal situation (large \( v \)), this reduces to \( H \thickapprox cT + RT \). This simplification helps in solving for molar heat capacities.
Partial Derivatives
Partial derivatives are used to measure how a function changes with respect to one variable while keeping other variables constant. They are crucial in thermodynamics for finding molar heat capacities.

For example, molar heat capacity at constant volume \( c_{V} \) is found using the partial derivative of internal energy \( u \) with respect to temperature at constant volume \( V \):

\[ c_{V} = \bigg(\frac{\frac{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{}}}}}}}}{u}{T}}_}{\bigg)}_V \bigg) \]

Given \( u = cT - \frac{a}{v} \), the partial derivative gives:

\[ c_{V} = c \]

Similarly, the molar heat capacity at constant pressure \( c_{P} \) involves the partial derivative of enthalpy \( H \) with respect to temperature at constant pressure \( P \):

\[ c_{P} = \bigg(\frac{\frac{\boldsymbol{\boldsymbol{H}{T}}}{}}{\bigg)}_^{P}_{\bigg)} \]

For the simplified enthalpy expression \( H \thickapprox cT + RT \), the partial derivative gives:

\[ c_{P} = c + R \]

Thus, partial derivatives allow us to precisely find how heat capacities change with temperature, essential for accurate thermodynamic calculations.

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Most popular questions from this chapter

A copper wire of length \(1.317 \mathrm{~m}\) and diameter \(3.26 \times 10^{-4} \mathrm{~m}\) is blackened and placed along the axis of an evacuated glass tube. The wire is connected to a battery, a rheostat, an ammeter, and a voltmeter, and the current is increased until, at the moment the wire is about to melt, the ammeter reads \(12.8 \mathrm{~A}\) and the voltmeter reads \(20.2 \mathrm{~V}\). Assuming that all the energy supplied was radiated and that the radiation from the glass tube is negligible, calculate the melting temperature of copper.

A container of volume \(V\) contains \(n\) moles of gas at high pressure. Connected to the container is a capillary tube through which the gas may leak slowly out to the atmosphere, where the pressure is \(P_{0}\). Surrounding the container and capillary is a water bath, in which is immersed an electrical resistor. The gas is allowed to leak slowly through the capillary into the atmosphere while electrical energy is dissipated in the resistor at such a rate that the temperature of the gas, the container, the capillary, and the water is kept equal to that of the outside air. Show that, after as much gas as possible has leaked out during time interval \(t\), the change in internal energy is $$ \Delta U=\mathcal{E} I t-P_{0}\left(n v_{0}-V\right) $$ where \(v_{0}\) is the molar volume of the gas at atmospheric pressure, \(\mathcal{S}\) is the potential difference across the resistor, and \(I\) is the current in the resistor.

A combustion cxperiment is performed by burning a mixture of fuel and oxygen in a constant-volume container surrounded by a water bath. During the experiment, the temperature of the water rises. If the system is the mixture of fuel and oxygen: (a) Has heat been transferred? (b) Has work been done? (c) What is the sign of \(\Delta U\) ?

(a) A small body with temperature \(T\) and emissivity \(\epsilon\) is placed in a large evacuated cavity with interior walls kept at temperature \(T_{W}\). When \(T_{W}-T\) is small, show that the rate of heat transfer by radiation is $$ \frac{\mathrm{d} Q}{d t}=4 T_{W}^{3} A \in \sigma\left(T_{W}-T\right) $$ (b) If the body remains at constant pressure, show that the time for the temperature of the body to change from \(T_{1}\) to \(T_{2}\) is given by $$ t=\frac{C_{P}}{4 T_{W}^{3} A \epsilon \sigma} \ln \frac{T_{W}-T_{1}}{T_{W}-T_{2}} $$ (c) Two small blackened spheres of identical size, one of copper and the other of aluminum, are suspended by silk threads within a large hole in a block of melting ice. It is found that it takes \(10 \mathrm{~min}\) for the temperature of the aluminum to drop from 276 to \(274 \mathrm{~K}\), and \(14.2\) min for the copper to drop the same interval of temperature. What is the ratio of specific heats of aluminum and copper? (The densities of \(\mathrm{Al}\) and \(\mathrm{Cu}\) are \(2.70 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and \(8.96 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) at \(25^{\circ} \mathrm{C}\), respectively.)

A container with rigid well-insulated walls is divided into two parts by a partition. One part contains a gas, and the other is evacuated. If the partition suddenly breaks, show that the initial and final internal energies of the gas are equal. (Note: this process is called an adiabatic free expansion.)

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