Advanced Chemistry - Mono Mole

February 7, 2023September 26, 2024

Carnot cycle

The Carnot cycle, developed by the French scientist Sadi Carnot in 1824, is a theoretical construct that sets the maximum efficiency of a thermodynamic engine.

Consider a gas in a piston-cylinder device (see diagram above) that repeatedly undergoes a cycle consisting of four reversible processes as follows:

(A to B) The pressure and temperature of the gas at A is equal to $p_{ext}$ and $T_h$ respectively. At this point, the device is in thermal contact with a hot reservoir at a constant temperature $T_h$ . Heat is transferred from the hot reservoir to the gas with $p_{ext}$ being reduced infinitesimally, causing the gas to expand and push the piston up to do work on the surroundings. This process is a reversible isothermal expansion of the gas.

(B to C) The contact with the hot reservoir is removed and the device is thermally insulated. $p_{ext}$ is again allowed to decrease infinitesimally and the gas continues to expand to do work on the surroundings with its internal energy, resulting in a drop in temperature to $T_c$ . This process is a reversible adiabatic expansion of the gas.

(C to D) The device is placed in contact with a cold reservoir at a constant temperature $T_c$ and the insulation is removed. Work is now done on the gas with $p_{ext}$ increased infinitesimally and heat is transferred from the gas to the reservoir. This process is a reversible isothermal compression of the gas.

(D to A) The contact with the cold reservoir is removed and the device is thermally insulated again. Work done on the gas continues with $p_{ext}$ increasing infinitesimally, resulting in an increase in its internal energy and a hence an increase in temperature back to $T_h$ . This process is a reversible adiabatic compression of the gas.

The theoretical piston-cylinder device that operates via the Carnot cycle is called a Carnot heat engine, i.e. one that operates by transferring energy from a body at a higher constant temperature to a body at a lower constant temperature and, through a cyclic process, converting some of that energy to mechanical work (see diagram below).

By the law on conservation of energy:

$-q_i=w_{cy}=q_o\;\;\;\;\;\;\;\;(95)$

where $q_i$ is the transfer of heat into the system from the hot reservoir ( $q_i>0$ ), $w_{cy}$ is the net work produced by the cycle ( $w_{cy}<0$ ) and $q_o$ is the transfer of thermal energy out of the system into the cold reservoir ( $q_o<0$ ).

The efficiency, $\varepsilon$ , of a heat engine is defined as the fraction of heat transferred to the system from the hot body that is converted to work:

$\varepsilon=\frac{\vert w_{cy}\vert}{q_i}\;\;\;\;\;\;\;\;(96)$

The modulus sign ensures that the value of $\varepsilon$ is positive. Substituting eq95 in eq96,

$\varepsilon=1+\frac{q_o}{q_i}\;\;\;\;\;\;\;\;(97)$

The greater the work produced by the engine, the smaller the value of $q_o$ and hence the closer the efficiency of the engine to one.

For a reversible heat engine to function, it can operate via different processes (e.g. reversible isochoric processes in place of reversible adiabatic processes). However, regardless of the processes involved, it must undergo a cycle where thermal energy is transferred from a hot reservoir to a cold reservoir with some of the energy converted to work, which is how a heat engine is defined. Therefore, it is impossible to construct an engine that undergoes a cyclic process where thermal energy extracted from the hot reservoir is completely converted to work without any energy deposited at the cold reservoir (see diagram above). This is the Kelvin-Planck statement of the second law of thermodynamics.

Question

Can an engine cycle consist of just the first and third processes i.e. a reversible isothermal expansion followed by a reversible isothermal compression?

Answer

If so, the net work done by the cycle is zero (net area under the PV curve is zero), which is not much of an engine.

An irreversible heat engine (real heat engine) is less efficient than a Carnot engine due to the occurrence of dissipative processes during the cycle, e.g. friction. Some of the work is converted to heat, which is transferred to the cold reservoir as $q_f$ (see diagram above). So,

$\vert w_{cy,irrev}\vert<\vert w_{cy,rev}\vert$

and according to eq96,

$\varepsilon_{irrev}<\varepsilon_{rev}$

Even though we have proven that an irreversible heat engine is less efficient than a Carnot engine, we cannot conclude that the Carnot cycle sets the maximum efficiency of a thermodynamic engine. To do so, we need to understand the Carnot heat pump and Carnot’s theorem.

Question

For an ideal gas,

$\oint dU=0\;\;\;\;\;\;\;\;(98)$

Show that eq98 is applicable to any reversible cycle for a system containing any substance.

Answer

Let’s assume that Carnot cycles of different working substances may have different shapes. Therefore, we need to prove that $\oint dU=0$ applies to any cycle.

Consider a cyclic process AEFGBCDA (see diagram above). $\oint dU$ for AEFGBCDA is:

$\oint dU_1=\int_A^EdU+\int_E^FdU+\int_F^GdU$
$+\int_G^BdU+\int_B^CdU+\int_C^DdU+\int_D^AdU\;\;\;\;\;\;\;\;(99)$

This cycle is actually made up of two Carnot cycles ABCDA and FGBEF with $\oint dU$ for ABCDA given by

$\oint dU_2=\int_A^BdU+\int_B^CdU+\int_C^DdU+\int_D^AdU\;\;\;\;\;\;\;\;(100)$

and $\oint dU$ for FGBEF being:

$\oint dU_3=\int_F^GdU+\int_G^BdU+\int_B^EdU+\int_E^FdU\;\;\;\;\;\;\;\;(101)$

Since $\int_E^BdU=-\int_B^EdU$ and $\int_A^EdU+\int_E^BdU=\int_A^BdU$ ,

$\int_A^EdU-\int_B^EdU=\int_A^BdU\;\;\;\;\;\;\;\;(102)$

Substitute eq102 in eq100

$\oint dU_2=\int_A^EdU-\int_B^EdU+\int_B^CdU+\int_C^DdU+\int_D^AdU\;\;\;\;\;\;\;\;(103)$

Adding eq101 and eq103 and comparing with eq99,

$\oint dU_1=\oint dU_2+\oint dU_3$

From eq98, $\oint dU_2=0$ and $\oint dU_3=0$ . Therefore, $\oint dU_1=0$ . This means that eq98 holds true for a cycle that is composed of two different sized Carnot cycles. The same logic can be applied to cycles that are made up of more than two Carnot cycles.

Let’s now look at an arbitrary reversible cycle that can represent a system containing any substance (see diagram above). The arbitrary cycle is exactly composed of an infinite number of Carnot cycles as the sizes of the Carnot cycles approach zero. Hence, for any arbitrary reversible cycle,

$\oint dU=\oint dU_{cyc1}+\oint dU_{cyc2}+\cdots+\oint dU_{cyc\infty}=0\;\;\;\;\;\;\;\;(105)$

Therefore, eq98 applies to any reversible cycle for a system containing any substance.

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February 7, 2023November 2, 2024

Second law of thermodynamics (overview)

The second law of thermodynamics states that the entropy of an isolated system increases during a spontaneous process. Another commonly used definition is the Kelvin-Planck statement, which we shall introduce in the next article.

The first law of thermodynamics states that the total energy of a system and its surroundings is constant. The state functions $U$ and $H$ , though very insightful and useful, are not able to indicate whether a process can proceed spontaneously. To determine the spontaneity of a process, scientists sought another state function. Rudolf Clausius, a German scientist, analysed the Carnot cycle, which had been developed earlier by Sadi Carnot, a French engineer, and eventually conceptualised a new thermodynamic state function $S$ , called entropy, which predicts the spontaneity of certain processes.

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February 7, 2023February 1, 2026

Polytropic processes

A polytropic process is described by the formula $pV^n=c$ , where $n\in\mathbb{R}_{\geq 0}$ . The table below shows the association of different pV curves for ideal gases with the corresponding thermodynamic processes:

Question

Why is the projection (the curve obtained by representing the process in $(p,V,T)$ space on the $p-V$ plane ) an adiabat when $n=\frac{C_p}{C_V}$ and why is $V$ a constant when $n=\infty$ ?

Answer

From eq44 and eq46, $dU=C_VdT$ and $dH=C_pdT$ for an ideal gas. For adiabatic processes, $dU=w=-pdV$ and hence, $C_VdT=-pdV$ . Furthermore, $H=U+PV\;\;\Rightarrow\;\;dH=dU+pdV+Vdp$ . Since $dU+pdV=0$ , $dH=Vdp\;\;\Rightarrow\;\;C_pdT=Vdp$ . So,

$\frac{C_p}{C_V}=-\frac{Vdp}{pdV}\;\;\Rightarrow\;\;\frac{C_p}{C_V}\int\frac{dV}{V}=-\int\frac{dp}{p}$
$\Rightarrow\;\;\frac{C_p}{C_V}(lnV+c_1)=-(lnp+c_2)$

which rearranges to $pV^{\frac{C_p}{C_V}}=c_3$ , where $c_3=e^{-c_2-\frac{C_p}{C_V}c_1}$ .

For the second why, $pV^n=c\;\;\Rightarrow\;\;V=\left(\frac{c}{p}\right)^{\frac{1}{n}}$ . If $n\rightarrow\infty$ , $V\rightarrow 1$ .

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February 7, 2023September 26, 2024

Line integral

A line integral is the integral of a multi-variable function along a curve. Line integrals can be categorised into scalar line integrals and vector line integrals. They are usually evaluated using parametric equations, which are then reduced to one or more scalar integrals.

Scalar line integrals

Let’s begin with scalar line integrals. A scalar function of two variables $f(x,y)$ is represented geometrically by a surface on a three-dimensional graph (see diagram above). Consider two points, A and B, on the surface. There are infinite ways to move along the surface between these two points. For simplicity, we have indicated two of these infinite paths using the red and pink curves, which can be projected onto the xy plane for better visualisation (see diagram below).

Due to the many paths available between the two points, the integral of $f(x,y)$ between the points is not well defined. In other words, the integral is path-dependent. For a particular path, the integral of $f(x,y)$ along this path, which is specified by one of the curves on the $xy$ plane, is called a line integral.

Geometrically, the result is the area of the ‘curtain’ extending from that path, between $(x_1,y_1)$ and $(x_2,y_2)$ on the xy plane, to the curve on the surface (see diagram above), i.e.:

$\int_cfds$

where $c$ denotes ‘curve’, which is the specified path linking the two points on the xy plane, and $ds$ is the infinitesimal change in arc length along the curve.

Since the specific path $c$ is also a function of $x$ and $y$ , we convert $f(x,y)$ and $c$ into parametric equations consisting of a single parameter and carry out the integration analytically (see 3rd Q&A below for an example).

Question

What if the two points, A and B, are the same?

Answer

In the scenario where the two points are the same, the infinite paths from $(x_1,y_1)$ to $(x_1,y_1)$ are infinite loops. Similarly, integrating the function via different loops may give different results. Each line integral in this case is represented by:

$\oint_cfds$

where the circle on the integral sign denotes a loop or a cyclic process.

Vector line integrals

Let’s now look at vector line integrals, an example of which is the work done by a vector field (see diagram above). For instance, the work done by a variable force $\boldsymbol{\mathit{F}}$ on a charged particle moving along some path in an electric field is

$w=\int_c\boldsymbol{\mathit{F}}\cdot d\boldsymbol{s}\;\;\;\;\;\;\;\;(92)$

As mentioned in the opening paragraph, such a line integral is evaluated by converting it into one or more scalar integrals. We write $\boldsymbol{\mathit{F}}$ and $d\boldsymbol{\mathit{s}}$ in terms of their two-dimensional Cartesian components:

$\int_c\boldsymbol{\mathit{F}}\cdot d\boldsymbol{s}=\int_c(F_x\boldsymbol{i}+F_y\boldsymbol{j})\cdot( dx\boldsymbol{i}+dy\boldsymbol{j})=\int_cF_xdx+\int_cF_ydy$

which can then be evaluated when $c$ is specified.

Question

What is the difference between a scalar field and a vector field?

Answer

A scalar field, expressed by a scalar function $f(x,y)$ , associates a scalar with each point in some region of space, while a vector field, expressed by a vector function $\boldsymbol{\mathit{F}}(x,y)$ associates a vector with each point.

Line integral with respect to coordinates

Line integrals can also be carried out with respect to one of the function variables instead of with respect to the arc length, e.g. the scalar integral $\int_cfdy$ . The geometric interpretation of $\int_cfdy$ is the projection of $\int_cfds$ on the $fy$ plane (see above diagram). An example of such a line integral is the work done on an ideal gas in a reversible process:

$w=-\int_cp(V,T)dV$

If the path of the above integral is defined by introducing a constraint, e.g. when $T=C$ , where $C$ is a constant, we get a plane that intersects with the surface at a particular contour (see diagram below). A single curve, an isothermal curve, is then projected on the $pV$ plane.

Consequently, the integral reduces to one involving a single variable:

$w=-\int_cp(V)dV=-nRT\int_{V_1}^{V_2}\frac{1}{V}dV$

Similarly, if the intersecting plane is $p=C$ or $V=C$ , the projection on the $pV$ plane is a horizontal line (isobaric process) or a vertical line (isochoric process) respectively.

Line integral of differential forms

Line integrals are sometimes written in differential form. Consider the work done by a vector field $\boldsymbol{\mathit{F}}(x,y)$ on a particle moving along some path $d\boldsymbol{s}(dx,dy)$ . We can write $\boldsymbol{\mathit{F}}$ and $d\boldsymbol{s}$ in terms of their components: $\boldsymbol{\mathit{F}}=F_x\boldsymbol{i}+F_y\boldsymbol{j}$ and $d\boldsymbol{s}=dx\boldsymbol{i}+dy\boldsymbol{j}$ . So, eq92 becomes

$w=\int_cdw=\int_c(F_x\boldsymbol{i}+F_y\boldsymbol{j}\cdot (dx\boldsymbol{i}+dy\boldsymbol{j})=\int_c\left(F_xdx+F_ydy\right)$

Compared to the general differential equation of two independent variables of the form $du=M(x,y)dx+N(x,y)dy$ , the RHS of the third equality of the above equation is the line integral of a differential equation, which in this case is an inexact differential. In other words, $\int_c\left(F_xdx+F_ydy\right)=\int_cdu$ .

In the case of an exact differential $df$ , its line integral is equal to the difference in values of the function $f$ at the final point and at the starting point:

$\int_cdf=f(x_2,y_2)-f(x_1,y_1)=\Delta f\;\;\;\;\;\;\;\;(93)$

This is known as the fundamental theorem of line integral. We call such a function, whose output is independent of the path taken to reach it, a state function.

Question

How do we proof the fundamental theorem of line integral?

Answer

Consider a function $f(x,y)$ with an exact differential of the form $df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$ or

$df=\nabla f\cdot d\boldsymbol{r}$

where $\nabla f=\left(\boldsymbol{i}\frac{\partial}{\partial x}+\boldsymbol{j}\frac{\partial}{\partial y}\right)f$ and $d\boldsymbol{r}=dx\boldsymbol{i}+dy\boldsymbol{j}$ .

To carry out the line integral between two points $A$ and $B$ , we convert the function into its parametric form, which is to let $x$ and $y$ be some function of $t$ , i.e. $f\left(x(t),y(t)\right)\equiv f(t)$ . Correspondingly, we have $\boldsymbol{r}=x(t)\boldsymbol{i}+y(t)\boldsymbol{j}$ and hence $\frac{d\boldsymbol{r}}{dt}=\frac{dx(t)}{dt}\boldsymbol{i}+\frac{dy(t)}{dt}\boldsymbol{j}$ . So,

$\int_cdf=\int_c\nabla f\cdot d\boldsymbol{r}=\int_c\left[\frac{\partial f}{\partial x(t)}\frac{dx(t)}{dt}dt+\frac{\partial f}{\partial y(t)}\frac{dy(t)}{dt}dt\right]$

$\int_cdf=\int_A^B\left[\frac{\partial f}{\partial x(t)}\frac{dx(t)}{dt}+\frac{\partial f}{\partial y(t)}\frac{dy(t)}{dt}\right]dt$

Using the definition of the chain rule for a multivariable function (see eq14b),

$\int_cdf=\int_A^B\frac{df}{dt}dt=f(B)-f(A)\equiv f\left(x(B),y(B)\right)-f\left(x(A),y(A)\right)=\Delta f$

Finally, we have shown in a previous article that the change in internal energy of a system $\Delta U$ is path-independent for a system containing a perfect gas. We will show in the article on entropy that $\Delta U$ is path-independent for any system. This makes $dU$ an exact differential and $U$ , a state function. Thus, we can write

$\int_cdU=U(x_2,y_2)-U(x_1,y_1)=\Delta U$

If the two points are the same, $(x_2,y_2)=(x_1,y_1)$ ,

$\oint_cdU=\Delta U=0\;\;\;\;\;\;\;\;(94)$

Therefore, the change in internal energy in a cyclic process is zero: $\oint_cdU=0$ . In general, the line integral of any exact differential involved in a cyclic process is zero.

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February 7, 2023February 13, 2025

Applications of the Joule-Thomson effect

Two significant applications of the Joule-Thomson effect involve the Hampson-Linde cycle and the vapour-compression cycle.

The Hampson-Linde cycle, also known as the ‘Linde refrigerator’ (see diagram below), employs the Joule-Thomson effect to liquefy gases.

The cycle is as follows:

1. The gas is compressed to allow it to circulate throughout the system. This raises the temperature of the gas because the kinetic energy of the gas molecules increases as they collide with the moving wall.
2. The gas cools as it passes through the heat exchanger and cools further when it expands through the Joule-Thomson orifice. The system is designed with specific pressures before and after the orifice to ensure the gas cools upon expansion, as described by the Joule-Thomson effect.
3. The low pressure gas is routed to the heat exchanger, cooling the warmer incoming gas, before returning to the compression chamber.
4. Warm replenishment gas for replacing any liquefied gas is mixed with the returning gas. The net temperature of the gas after mixing is lower than its temperature in step1. Steps 1 to 4 is repeated until the air is cool enough to condense into the compartment below the orifice.

The vapour-compression cycle is used in household refrigerators and air conditioning units.

The cycle is as follows:

1. Compression: The refrigerant gas is compressed by the compressor, allowing it to circulate throughout the system. This is not an isenthalpic process because work is done on the gas, which increases both the pressure and temperature of the refrigerant.
2. Condensation: The hot, high-pressure refrigerant gas then passes through the condenser coils, usually located on the back or bottom of the appliance. As it cools, it condenses into a high-pressure liquid.
3. Expansion: The flow of the high-pressure liquid refrigerant through an expansion valve, known as the Joule-Thomson orifice, can be modelled using the Joule-Thomson experiment. This causes the refrigerant’s pressure to drop, resulting in a mixture of liquid and vapour. The refrigerant is carefully selected based on its inversion temperature to ensure it cools upon expansion, as described by the Joule-Thomson effect.
4. Evaporation: The low-pressure refrigerant then enters the evaporator coils (inside the refrigerator or air handler of an air conditioning unit). Here, it absorbs heat from the surrounding environment, which allows it to fully evaporate into a gas. This process cools down the air inside the refrigerator or the room.
5. Cycle repeats: The refrigerant gas is then drawn back into the compressor to repeat the cycle. The desired temperature of the room or the interior of the refrigerator is monitored by a thermostat, which then signals the appropriate compression pressure to apply for subsequent cycles.

The desired temperature of the room or the interior of the refrigerator is monitored by a thermostat, which then signals the appropriate compression pressure to apply for subsequent cycles.

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February 7, 2023March 24, 2025

The Joule-Thomson experiment

The Joule-Thomson experiment, an improved version of the Joule experiment, was conducted by James Joule and William Thomson in 1852 to study the thermodynamic properties of a gas expanding into a vacuum.

The experiment involves a system that is adiabatically insulated from its surroundings (see diagram above). The system consists of a double-piston cylinder that is separated into two compartments by a rigid porous plug. Constant pressures $p_1$ and $p_2$ , where $p_1>p_2$ , are applied to the left piston and the right piston respectively. As a result, the gas in the left compartment is slowly throttled irreversibly to the right compartment. Assuming that pressure is well defined in each compartment, $w_1=-\int_{V_1}^0p_1dV=p_1V_1$ and $w_2=-\int_{0}^{V_2}p_2dV=-p_2V_2$ . The total work done on the system is

$w=w_1+w_2=p_1V_1-p_2V_2$

According to the first law of thermodynamics for the entire system,

$\Delta U=U_2-U_1=q+w=p_1V_1-p_2V_2$

$U_2+p_2V_2=U_1+p_2V_2$

So, $H_2=H_1$ or $\Delta H=0$ . If there is a change in temperature of the system, the experiment measures the change in gas temperature with the change in gas pressure at constant enthalpy, i.e. $\left(\frac{\partial T}{\partial p}\right)_H$ , which is defined as the Joule-Thomson coefficient $\mu_{JT}$ :

$\mu_{JT}=\left(\frac{\partial T}{\partial p}\right)_H\;\;\;\;\;\;\;\;(90)$

The experiment is conducted multiple times by lowering $p_2$ for a constant $p_1$ and $T_1$ . A temperature-pressure plot of the results gives an isenthalpic curve (see diagram above). The gradient at a particular point of the curve is the Joule-Thomson coefficient with respect to $T$ and $p$ at that point. Since the change in pressure of the system is always negative, points on the curve that are associated with negative gradients correspond to the gas warming on expansion. Conversely, points on the curve that are associated with positive gradients correspond to the gas cooling on expansion. The point at which the gradient is zero is called the inversion point. Repeating the experiment by holding $T_1$ at new constant values and varying $p_2$ , we have multiple isenthapic curves:

The line connecting all inversion points is the inversion curve. The area on the left of the inversion curve is where cooling occurs for a gas when it expands into a vacuum, while the area on the right is where heating occurs. The temperature corresponding to the inversion curve at a given pressure is the inversion temperature of the gas at that pressure. If a vertical line is extended across the inversion curve at a particular pressure, it will intersect the curve at two points: an upper inversion temperature and a lower inversion temperature. Cooling upon expansion usually occurs when the temperature of the gas is between its upper and lower inversion temperatures prior to expansion. In other words, whether the expansion results in heating or cooling of the gas depends on the initial conditions and the properties of the gas. This phenomenon, known as the Joule-Thomson effect, is exploited in refrigeration processes.

Modern methods of measuring $\mu_{JT}$ involve evaluating $\left(\frac{\partial H}{\partial p}\right)_T$ , which is known as the isothermal Joule-Thomson coefficient. To determine the relation between $\mu_{JT}$ and $\left(\frac{\partial H}{\partial p}\right)_T$ , we refer to eq16, where $\left(\frac{\partial H}{\partial p}\right)_T=-\frac{1}{\left(\frac{\partial T}{\partial H}\right)_p\left(\frac{\partial p}{\partial T}\right)_H}$ . Using eq15, $\left(\frac{\partial H}{\partial p}\right)_T=-\left(\frac{\partial H}{\partial T}\right)_p\left(\frac{\partial T}{\partial p}\right)_H$ . Substituting eq46 and eq90 in this expression,

$\left(\frac{\partial H}{\partial p}\right)_T=-C_p\mu_{JT}\;\;\;\;\;\;\;\;(91)$

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February 7, 2023September 26, 2024

The Joule experiment

The Joule experiment was an attempt by James Joule in 1843 to determine $\left(\frac{\partial U}{\partial V}\right)_T$ for a gas expanding into a vacuum.

The experiment involves a system that consists of a gas-filled compartment (A) and an evacuated compartment (B), both of which are immersed in a water bath (see diagram above). When the valve between the compartments is opened, the gas expands irreversibly from A to B, doing no work in the process. Assuming that no heat transfer occurs between the system and its surroundings, the first law of thermodynamics states that $\Delta U=0$ . If there is a change in temperature of the water bath, the experiment measures the change in gas temperature with the change in gas volume at constant internal energy, i.e. $\left(\frac{\partial T}{\partial V}\right)_U$ , which is defined as the Joule coefficient $\mu_J$ :

$\mu_J=\left(\frac{\partial T}{\partial V}\right)_U\;\;\;\;\;\;\;\;(88)$

To determine $\left(\frac{\partial U}{\partial V}\right)_T$ , we refer to eq16, where $\left(\frac{\partial U}{\partial V}\right)_T=-\frac{1}{\left(\frac{\partial T}{\partial U}\right)_V\left(\frac{\partial V}{\partial T}\right)_U}$ . Using eq15, $\left(\frac{\partial U}{\partial V}\right)_T=-\left(\frac{\partial U}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_U$ . Substituting eq44 and eq88 in this expression,

$\left(\frac{\partial U}{\partial V}\right)_T=-C_V\mu_J\;\;\;\;\;\;\;\;(89)$

For an ideal gas, $U$ is independent of the volume of the gas, i.e. $\left(\frac{\partial U}{\partial V}\right)_T=0$ , because there is no intermolecular forces between the molecules (see eq43a). We would therefore expect a non-zero $\left(\frac{\partial U}{\partial V}\right)_T$ for a real gas. However, Joule’s experiment setup was too crude to measure a non-zero $\mu_J$ . An improved version of Joule’s experiment is the Joule-Thomson experiment.

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February 7, 2023August 25, 2025

Reversible adiabatic processes

A reversible adiabatic process is a reversible thermodynamic process in which no heat or mass transfer occurs.

A reversible adiabatic expansion follows the path from B to C, while a reversible adiabatic compression follows the path from D to A. The curves describing the paths of adiabatic processes are called adiabats (see diagram above, where we have included two isotherms, AB and CD, for comparison).

A reversible adiabatic compression is carried out in a frictionless, insulated piston-cylinder device containing a gas (see diagram above). The gas undergoes infinitesimal steps of compression, resulting in an increase in the internal energy of the system and therefore the temperature of the system. Since no heat enters or leaves the system, $\delta q=0$ and according to the first law of thermodynamics,

$dU=\delta w$

For an ideal gas, we have shown in a previous article that $dU=C_V(T)dT$ and hence, work done on the ideal gas system for a reversible adiabatic compression is:

$\delta w=C_V(T)dT\;\;\;\;\;\;\;\;(84)$

Substituting $\delta w=\frac{nRT}{V}dV$ in eq84 and integrating throughout gives:

$nR\int_{V_D}^{V_A}\frac{1}{V}dV=C_V\int_{T_c}^{T_h}\frac{1}{T}dT$

where we have assumed that the heat capacity of the ideal gas is independent of temperature over the temperature range of interest and therefore a constant.

Solving the above equation yields:

$\frac{V_A}{V_D}=\left(\frac{T_h}{T_c}\right)^{\alpha}\;\;\;\;\;\;\;\;(85)$

where $\alpha=\frac{C_V}{nR}$ .

Similarly, for an adiabatic expansion from B to C, we have

$\frac{V_C}{V_B}=\left(\frac{T_c}{T_h}\right)^{\alpha}\;\;\;\;\;\;\;\;(86)$

Equating eq85 and eq86,

$\frac{V_A}{V_D}=\frac{V_B}{V_C}\;\;\;\;\;\;\;\;(87)$

Question

Show that internal energy of a system $U$ containing a perfect gas is a state function using the diagram above where

AC: reversible isothermal process
AD: reversible adiabatic process
AB: reversible isobaric process
BC & CD: reversible isochoric processes

Answer

Consider the change in internal energy of a system containing one mole of ideal gas from point A to point C. The change can be brought about by three different paths, with the first being a reversible isothermal expansion (AC), where $\Delta U_{AC}=0$ .

The second path involves a reversible adiabatic expansion (AD) followed by a reversible isochoric heating (DC). The changes in internal energy for paths AD and DC are:

$\Delta U_{AD}=q_{AD}+w_{AD}=0+\int_{T_C}^{T_D}C_{V,m}dT$

$\Delta U_{DC}=q_{DC}+w_{DC}=\int_{T_D}^{T_C}C_{V,m}dT+0$

The change in internal energy for path ADC is:

$\Delta U_{ADC}=\Delta U_{DC}+\Delta U_{AD}$

Since $\int_{T_C}^{T_D}C_{V,m}dT=-\int_{T_D}^{T_C}C_{V,m}dT$

$\Delta U_{ADC}=0$

The third path consists of a reversible isobaric expansion (AB) followed by a reversible isochoric cooling (BC). The change in internal energy for path AB is:

$\Delta U_{AB}=q_{AB}+w_{AB}=\int_{T_A}^{T_B}C_{p,m}dT-\int_{V_1}^{V_2}p_3dV$

From a previous article, $C_{p,m}-C_{V,m}=R$ , and so

$\Delta U_{AB}=R(T_B-T_A)+\int_{T_A}^{T_B}C_{V,m}dT-p_3(V_2-V_1)$

Substituting $V_2=\frac{RT_B}{p_3}$ and $V_1=\frac{RT_A}{p_3}$ in the above equation,

$\Delta U_{AB}=\int_{T_A}^{T_B}C_{V,m}dT=\int_{T_C}^{T_B}C_{V,m}dT$

The change in internal energy for path BC is:

$\Delta U_{BC}=q_{BC}+w_{BC}=\int_{T_B}^{T_C}C_{V,m}dT+0$

The change in internal energy for path ABC is:

$\Delta U_{ABC}=\Delta U_{BC}+\Delta U_{AB}=0$

Regardless of the path taken, the change in internal energy from point A to point C is the same. Therefore, $U$ is a state function.

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Using Kirchhoff’s law to determine the standard enthalpy change of ionisation and electron gain

Ionisation energy of a species is defined as the energy for removing an electron from the ground state of the species. According to statistical thermodynamics, the ground state of a species is the electronic configuration of that species at absolute zero. If ionisation energy is defined at any other temperatures above absolute zero, there will be a range of ionisation energies for a particular species, as the electron then is removed from various excited states. Hence, ionisation energy is equal to the difference between the sum of enthalpies of formation of the ionised species and the electron at absolute zero, and the enthalpy of formation of the species at absolute zero.

$M(g)\rightarrow M^+(g)+e^-$
$\Delta H_r^{\circ}(0)=\Delta H_{f,M^+}^{\circ}(0)+\Delta H_{f,e^-}^{\circ}(0)-\Delta H_{f,M}^{\circ}(0)=IE$

Ionisation energies are presented in data tables using eq73 of the previous article, where $T_1$ is the experimental temperature and $T_2$ is absolute zero.

Question

Calculate the first ionisation enthalpy of magnesium at 298.15K, given IE₁ = 738 kJmol^-1 and assuming the heat capacities at constant pressure of all species (including electron) are given by that of a perfect monoatomic gas of $\frac{5}{2}R$ .

Answer

$Mg(g)\rightarrow Mg^+(g)+e^-$
$\Delta H_r^{\circ}(0)=+738\;kJmol^{-1}$
Using eq73,

$\Delta H_r^{\circ}(298.15)-738000=\int_{0}^{298.15}\frac{5}{2}RdT+\int_{0}^{298.15}\frac{5}{2}RdT-\int_{0}^{298.15}\frac{5}{2}RdT$

$\Delta H_r^{\circ}(298.15)=+744\;kJmol^{-1}$

With reference to the above example where the heat capacities at constant pressure of all species (including electron) are given by that of a perfect monoatomic gas of $\frac{5}{2}R$ ,

$\Delta H_{1st\;ion}^{\circ}(T)-\Delta H_r^{\circ}(0)=\int_{0}^{T}\frac{5}{2}RdT+\int_{0}^{T}\frac{5}{2}RdT-\int_{0}^{T}\frac{5}{2}RdT=\int_{0}^{T}\frac{5}{2}RdT$

$\Delta H_{1st\;ion}^{\circ}(T)=\Delta H_r^{\circ}(0)+\frac{5}{2}RT=IE_1+\frac{5}{2}RT\;\;\;\;\;\;\;\;(80)$

where $IE_1$ is at absolute zero.

Similarly,

$\Delta H_{2nd\;ion}^{\circ}(T)=IE_2+\frac{5}{2}RT\;\;\;\;\;\;\;\;(81)$

Therefore, the ionisation enthalpy of $M(g)\rightarrow M^{2+}(g)+2e^-$ is

$\Delta H_{total\;ion}^{\circ}(T)=IE_1+IE_2+5RT\;\;\;\;\;\;\;\;(82)$

Electron affinity is also a zero Kelvin process, as it is defined as the energy for the addition of an electron to a species in its ground state. Although an exothermic process, electron affinity (EA) is quoted in data tables as positive values, i.e.

$M(g)+e^-\rightarrow M^-(g)\;\;\;\;\;\;\;\;\Delta H_r^{\circ}(0)=-EA$

Since the standard enthalpy change of electron gain of a species $X^{n-1}$ is the negative of the standard enthalpy change of ionisation of that species with an additional electron attached $X^n$ , i.e.

$X^n(g)\begin{matrix} IE\\\rightarrow \\ \leftarrow \\ EA \end{matrix} X^{n-1}(g)+e^-$

$\Delta H_{eg}^{\circ}[X^n]=-\Delta H_{ion}^{\circ}[X^{n-1}]\;\;\;\;\;\;\;\;(83)$

the enthalpy change of electron gain, assuming the heat capacities at constant pressure of all species (including electron) are given by that of a perfect monoatomic gas of $\frac{5}{2}R$ , is determined by substituting eq83 in eq80 and eq81:

$\Delta H_{1st\;eg}^{\circ}(T)=-\left[\Delta H_{r}^{\circ}(0)+\frac{5}{2}RT\right]=-EA_1-\frac{5}{2}RT$

$\Delta H_{2nd\;eg}^{\circ}(T)=-\left[\Delta H_{r}^{\circ}(0)+\frac{5}{2}RT\right]=-EA_2-\frac{5}{2}RT$

with the electron gain enthalpy of $M(g)+2e^-\rightarrow M^{2-}(g)$ being:

$\Delta H_{total\;eg}^{\circ}(T)=-EA_1-EA_2-5RT$

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Using Kirchhoff’s law to determine the standard enthalpy change of reaction at 298K

The experimental result of the standard change in enthalpy for the combustion reaction $CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$ is 281.93 kJmol^-1 at 1200K. Find its standard change in enthalpy at 298.15K given:

According to Hess’ law,

$\Delta H_r^{\circ}=\sum_Pv_P(\Delta H_f^{\circ})_P-\sum_Rv_R(\Delta H_f^{\circ})_R\;\;\;\;\;\;\;\;(69)$

At 1200K,

$281.93=\Delta H_{f,1200K}^{\circ}[CO_2(g)]-\Delta H_{f,1200K}^{\circ}[CO(g)]-\frac{1}{2}\Delta H_{f,1200K}^{\circ}[O_2(g)]\;\;\;\;\;\;\;\;(70)$

At 298.15K,

$\Delta H_{r,298.15K}^{\circ}=\Delta H_{f,298.15K}^{\circ}[CO_2(g)]-\Delta H_{f,298.15K}^{\circ}[CO(g)]-\frac{1}{2}\Delta H_{f,298.15K}^{\circ}[O_2(g)]\;\;\;\;\;\;\;\;(71)$

Subtracting eq71 from eq70,

$\begin{align}281.93-\Delta H_{r,298.15K}^{\circ}=&\Delta H_{f,1200K}^{\circ}[CO_2(g)]-\Delta H_{f,298.15K}^{\circ}[CO_2(g)]\\&-\{\Delta H_{f,1200K}^{\circ}[CO(g)]-\Delta H_{f,298.15K}^{\circ}[CO(g)]\}\\&-\frac{1}{2}\{\Delta H_{f,1200K}^{\circ}[O_2(g)]-\Delta H_{f,298.15K}^{\circ}[O_2(g)]\}\;\;\;\;\;\;\;\;(72)\end{align}$

Since all species in the reaction are in the gaseous state from 298.15K to 1200K, we can substitute eq66 in eq72,

$281.93-\Delta H_{r,298.15K}^{\circ}=\int_{1200}^{298.15}C_{p,CO_2}(T)dT-\int_{1200}^{298.15}C_{p,CO}(T)dT-\frac{1}{2}\int_{1200}^{298.15}C_{p,O_2}(T)dT\;\;\;\;\;\;\;\;(73)$

Substituting eq68 in the 1^st integral on the RHS of eq73,

$\int_{1200}^{298.15}C_{p,CO_2}(T)dT=\int_{1200}^{298.15}(a+bT+cT^2+dT^3)dT\;\;\;\;\;\;\;\;(74)$

Substituting the relevant values from the table above in eq73 and evaluating the integral (using the form expressed by eq74 for each integral in eq73),

$\int_{1200}^{298.15}C_{p,CO_2}(T)dT=-44.467\;kJmol^{-1}\;\;\;\;\;\;\;\;(75)$

$\int_{1200}^{298.15}C_{p,CO}(T)dT=-28.428\;kJmol^{-1}\;\;\;\;\;\;\;\;(76)$

$\frac{1}{2}\int_{1200}^{298.15}C_{p,O_2}(T)dT=-14.981\;kJmol^{-1}\;\;\;\;\;\;\;\;(77)$

and

$\Delta H_{r,298.15K}^{\circ}=+282.99\;kJmol^{-1}\;\;\;\;\;\;\;\;(78)$

Combining eq66 and eq69, for a reaction $A+B+\cdots\rightarrow M+N+\cdots$

$\Delta H_{r}^{\circ}(T_2)-\Delta H_{r}^{\circ}(T_1)=\sum_Pv_P\left[\int_{T_1}^{T_2}C_P(T)dT\right]_P-\sum_Rv_R\left[\int_{T_1}^{T_2}C_P(T)dT\right]_R\;\;\;\;\;\;\;\;(79)$

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