Advanced Chemistry - Mono Mole

February 7, 2023October 20, 2025

First law of thermodynamics

The first law of thermodynamics states that the change in internal energy of a system is the sum of the energy transferred to the system, due to a difference in temperature between the system and its surroundings, and the work done on the system by the surroundings.

The internal energy $U$ of a system is defined as the sum of all the energies of molecules in the system. These energies include:

Relativistic rest mass energies of electrons and nuclei (i.e. energies attributed to the existence of the molecules)
Intramolecular translational, rotational, vibrational and electronic energies
Intermolecular forces of interaction

$U$ is a thermodynamic property of a system. However, judging from the composition of $U$ , it is very difficult to measure $U$ directly. An alternate representation of $U$ therefore needs to be established to characterise the internal energy of a system.

According to the theory of conservation of energy, energy can neither be created nor destroyed but can be transformed from one form to another. Energy can also be transferred between thermodynamic systems or from a system to its surroundings.

Consider a closed piston-cylinder system containing a gas (see diagram above). The internal energy of the system increases when thermal energy flows into it, or when a force is applied on the piston, compressing the gas. By the theory of conservation of energy,

$\Delta U=q+w\;\;\;\;\;\;\;\;(24)$

where

$\Delta U$ is the change in internal energy of the system
$q$ (or heat) is the transfer of energy to the system due to a difference in temperature
$w$ (or work) is work done on the system by the surroundings (transfer of energy to the system not due to a difference in temperature)

Eq24 is known as the first law of thermodynamics. Instead of considering the absolute internal energy of a system when we study processes, we analyse the change in internal energy of the system, which is easily quantified by accounting for the amount of energy transferred to and from the system in the form of $w$ and $q$ (refer to the next article for quantification of $q$ ). From an earlier article, reversible work done on the system is given by $w=-\int_{V_i}^{V_f}pdV$ . Therefore, for a system undergoing a reversible PV process, eq24 becomes

$\Delta U=q-\int_{V_i}^{V_f}pdV\;\;\;\;\;\;\;\;(25)$

If the walls of the cylinder and piston are adiabatic, the system is isolated and $q=0$ . Eq24 becomes

$\Delta U=w\;\;\;\;\;\;\;\;(26)$

which states that work done on an isolated system is exactly equal to the increase in the internal energy of the system.

Question

Describe the work-energy theorem and show that it is consistent with eq26 for work done on a monoatomic ideal gas in an adiabatic piston-cylinder system.

Answer

The work-energy theorem states that the net work done on a particle is equal to the change in the kinetic energy of the particle. This can be shown as follows:

$w_{net}=\int_{x_i}^{x_f}F_{net,x}dx+\int_{y_i}^{y_f}F_{net,y}dy+\int_{z_i}^{z_f}F_{net,z}dz$

$w_{net}=\int_{x_i}^{x_f}m\frac{dv_x}{dt}dx+\int_{y_i}^{y_f}m\frac{dv_y}{dt}dy+\int_{z_i}^{z_f}m\frac{dv_z}{dt}dz$

$w_{net}=m\left(\int_{x_i}^{x_f}v_xdv_x+\int_{y_i}^{y_f}v_ydv_y+\int_{z_i}^{z_f}v_zdv_z\right)$

$w_{net}=\frac{1}{2}m\left(v_{xf}^2+v_{yf}^2+v_{zf}^2\right)-\frac{1}{2}m\left(v_{xi}^2+v_{yi}^2+v_{zi}^2\right)=\Delta KE$

Since the internal energy of a monoatomic ideal gas is entirely in the form of the kinetic energy of the gas, the work done on the gas in an adiabatic system is equal to the change in the internal energy of the gas.

Suppose the piston is now rendered immovable and the walls of the cylinder are thermally conducting, we have a closed constant volume system. Eq25 becomes

$\Delta U=q\;\;\;\;\;\;\;\;(27)$

The change in internal energy of the system at constant volume can then be experimentally determined by quantifying the heat transferred to the system.

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February 7, 2023November 24, 2025

Zeroth law of thermodynamics

The zeroth law of thermodynamics states that two thermodynamic systems, each in thermal equilibrium with a third, are in thermal equilibrium with each other.

Consider two systems, A and B, in contact with each other via an immovable thermally conducting boundary, achieving thermal equilibrium over time.

B is then brought in contact with a third system, C, via the same type of boundary.

If no net energy and matter transfer is observed between B and C, we can deduce that A is also in thermal equilibrium with C. When this happens, there must be at least one measurable thermodynamic property that is common among the three systems.

We call this common thermodynamic property, temperature, and the above-mentioned principle that allows us to define temperature, the zeroth law of thermodynamics. Temperature is therefore defined as the thermodynamic property that has the same magnitude in two systems that are in thermal equilibrium.

This conceptual framework has a long history. The principle that eventually became the zeroth law of thermodynamics was first mentioned in 1871 by James Maxwell, who said, “bodies whose temperatures are equal to that of the same body have themselves equal temperatures”. A modern version of the zeroth law of thermodynamics utilises the term ‘thermal equilibrium’ and states that

Two thermodynamic systems, each in thermal equilibrium with a third, are in thermal equilibrium with each other.

Question

Which of the three systems described above is the thermometer?

Answer

B, as it indicates that A and C have the same temperature. If B contains a substance with relatively high coefficient of thermal expansion, e.g., helium, it can be used to assess A and C at various temperatures by monitoring its volume at constant pressure. Thus, the zeroth law of thermodynamics forms the basis of thermometry.

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February 7, 2023September 26, 2024

Path function

A path function is a mathematical function that describes thermodynamic processes that are involved in a change of equilibrium states. Unlike a state function, whose output is independent of the path taken to reach it, the output of a path function is path-dependent.

An example of a path function is the work done by an ideal gas. Consider an ideal gas in a piston-cylinder device immersed in a water bath (see diagram below).

It is evident from the above PV diagram that there are three different paths for work done in bringing the system from one equilibrium state $(p_2,V_1,T)$ to another equilibrium state $(p_1,V_2,T)$ :

Amongst the three paths, A to B to C requires the greatest work, while A to D to C requires the least amount of work. This is because the system is expanding reversibly against a higher constant pressure throughout the process from A to B at $p_2$ , versus that from D to C at $p_1$ . The path from A to C requires intermediate work done as the system is expanding reversibly against a decreasing pressure from $p_2$ to $p_1$ . Relative work done by the three paths can be inspected visually by estimating the areas under the respective PV curves. The precise values are calculated using eq6 and eq7. Since work done has values that are dependent on the path taken from one thermodynamic state to another, it is a path function.

Lastly, the differential form of the reversible isobaric expansion of an ideal gas is

$\delta w=p_{ex}dV\;\;\;\;\;\;\;\;(23)$

We have used the symbol $\delta$ to emphasise that $w$ is a path function but the symbol $d$ is also acceptable. Eq23 is called an inexact differential because its integral is not path independent. Even though work has the same units (Joules) as energy, it is misleading to say that work is a form of energy, which if it is, will be described by a thermodynamic function that is path-independent. Work is rather, a process of energy transfer between a system and its surroundings.

Question

Show that $dq=C_V(T)dT+\frac{nRT}{V}dV$ is an inexact differential, while $\frac{dq}{T}$ is an exact differential.

Answer

Comparing the first equation to the general form of a differential equation $du=Mdx+Ndy$ , we have $M=C_V(T)$ and $N=\frac{nRT}{V}$ . We need to show that $\frac{\partial M}{\partial V}\neq\frac{\partial N}{\partial T}$ for an inexact differential:

$\frac{\partial}{\partial V}C_V(T)=0$

and

$\frac{\partial}{\partial T}\frac{nRT}{V}=\frac{nR}{V}$

For $\frac{dq}{T}$ , its second cross partial derivatives are $\frac{\partial}{\partial V}\left[\frac{C_V(T)}{T}\right]=0$ and $\frac{\partial}{\partial T}\left(\frac{nR}{V}\right)=0$ . Therefore $\frac{dq}{T}$ is an exact differential.

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February 7, 2023December 26, 2024

State function

A state function is a mathematical function that describes the thermodynamic state of a system at equilibrium.

The ideal gas equation is a state function because it describes the state of a gas at equilibrium with a specific set of values: volume, temperature and pressure. In other words, a state function for a system at equilibrium relates one or more input thermodynamic properties to an output thermodynamic property.

Since a particular state of a system at equilibrium is characterised by a set of numbers, the output value of a state function for that state is independent on the path (i.e. process) taken to reach that value.

Question

Show that the output of the state function $V(p,T)$ is independent on the path taken to reach it.

Answer

The total differential of $V$ is

$dV=\left(\frac{\partial V}{\partial T}\right)_pdT+\left(\frac{\partial V}{\partial p}\right)_Tdp\;\;\;\;\;\;\;\;(18)$

The second cross partial derivatives of $V$ are

$\frac{\partial}{\partial p}\left(\frac{\partial V}{\partial T}\right)\;\;\;\;\;\;\;\;(19)$

and

$\frac{\partial}{\partial T}\left(\frac{\partial V}{\partial p}\right)\;\;\;\;\;\;\;\;(20)$

Eq19 refers to the path where the function V is changed with respect T at constant p, followed by a change with respect to p at constant T, whereas eq20 refers to another path where the function V is changed with respect to p at constant T, followed by a change with respect to T at constant p. If eq19 is equal to eq20, the change of V is independent of the path taken. Substituting $V=\frac{nRT}{p}$ in eq19 and eq20 gives $-\frac{nR}{p^2}$ for both equations. Therefore, the change of V is path-independent and the differential given by eq18 is called an exact differential.

Finally, if the output value of the function of pressure and the output value of the function of volume are path-independent, then the output value of the product of the two functions or the output value of the sum of the two functions are also path-independent.

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February 7, 2023November 27, 2024

Exact and inexact differentials

An exact differential is a differential equation $du$ , for instance of two variables, of the form $du=M(x,y)dx+N(x,y)dy$ , where $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

Consider the function $u=xy+c$ , where $c$ is a constant. Its total differential is $du=\left(\frac{\partial u}{\partial x}\right)_ydx+\left(\frac{\partial u}{\partial y}\right)_xdy$ . Comparing with the general form of an exact differential, we have $M(x,y)=\left(\frac{\partial u}{\partial x}\right)_y$ and $N(x,y)=\left(\frac{\partial u}{\partial y}\right)_x$ . Since $\frac{\partial M}{\partial y}=\frac{\partial^2u}{\partial y\partial x}=1=\frac{\partial^2u}{\partial x\partial y}=\frac{\partial N}{\partial x}$ , $du$ is an exact differential. The equality of the mixed partials $\frac{\partial^2u}{\partial y\partial x}=\frac{\partial^2u}{\partial x\partial y}$ implies that the change in $u$ is independent of the path taken.

Conversely, an inexact differential is a differential equation $du$ of the form $du=M(x,y)dx+N(x,y)dy$ , where $\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$ . The change in $u$ , in this case, is dependent on the path taken.

Question

Show that $du=3ydx+xdy$ is an inexact differential.

Answer

We have $\frac{\partial M}{\partial y}=3$ and $\frac{\partial N}{\partial x}=1$ . So, $\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$ .

Another difference between an exact differential and an inexact differential is that an exact differential integrates directly to give the function $u$ , whereas an inexact differential does not.

Question

Using the integral criterion, show that $du=xdy+ydx$ is an exact differential, while $du=3ydx+xdy$ is an inexact differential.

Answer

The integrated form of the first differential is evidently $u=xy+c$ . The detailed analysis involves two steps. First, integrating $M(x,y)=\left(\frac{\partial u}{\partial x}\right)_y$ with respect to $x$ , treating $y$ as a constant, yields

$u=xy+f(y)\;\;\;\;\;\;\;\;17a$

The function $f(y)$ accounts for terms in $u$ involving $y$ or constants, which differentiate to zero when differentiating $u$ with respect to $x$ . Secondly, integrating $N(x,y)=\left(\frac{\partial u}{\partial y}\right)_x$ with respect to $y$ , with $x$ as a constant, gives

$u=xy+g(x)\;\;\;\;\;\;\;\;17b$

Comparing eq17a and eq17b, we find $u=xy+c$ , where $f(x)=cx^0=cy^0=f(y)=c$ . In other words, $u=xy+c$ integrates directly from $du=xdy+ydx$ .

For the second differential, integrating with respect to $x$ yields $u=3xy+g(y)$ , and integrating with respect to $y$ gives $u=xy+h(x)$ . Clearly, $3xy+g(y)\neq xy+h(x)$ , for all $g(y)$ and $h(x)$ , indicating that the second differential does not directly integrate to give a function.

The fact that an exact differential integrates directly to give the function $u$ but an inexact differential does not, implies that $M(x,y)$ and $N(x,y)$ for a differentiable function must be of the appropriate forms of $\left(\frac{\partial u}{\partial x}\right)_y$ and $\left(\frac{\partial u}{\partial y}\right)_x$ , respectively. In other words, the total differential of a differentiable function must be an exact differential.

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February 7, 2023September 26, 2024

Partial derivatives, the total differential and the multivariable chain rule

A partial derivative of a multi-variable function is its derivative with respect to one of those variables, with the other variables held constant. For example, the partial derivatives of $z=f(x,y)$ with respect to $x$ and $y$ are defined as

$\left(\frac{\partial z}{\partial x}\right)_y=\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}\;\;\;\;\;\;\;\;(12)$

$\left(\frac{\partial z}{\partial y}\right)_x=\lim_{\Delta y\rightarrow 0}\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\;\;\;\;\;\;\;\;(13)$

respectively, where the symbol $(\;\;)_i$ means that the variable $i$ is held constant (the symbol may be omitted for simplicity).

If $z=x^3y^2+e^{xy}$ , then $\left(\frac{\partial z}{\partial x}\right)_y=3x^2y^2+ye^{xy}$ .

The total differential of a multi-variable function is its change with respect to the changes in all the independent variables. For example, the total differential of the function $z=f(x,y)$ is

$dz=\left(\frac{\partial z}{\partial x}\right)_ydx+\left(\frac{\partial z}{\partial y}\right)_xdy\;\;\;\;\;\;\;\;(14)$

Question

How is eq14 derived?

Answer

The total change in $z$ is $\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y)$ , which is equivalent to

$\Delta z=f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)+f(x,y+\Delta y)-f(x,y)$

Multiplying the 1^st and 2^nd terms on the RHS of the above equation by $\frac{\Delta x}{\Delta x}$ and the 3^rd and 4^th terms by $\frac{\Delta y}{\Delta y}$ ,

$\Delta z=\left[\frac{f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)}{\Delta x}\right]\Delta x+\left[\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\right]\Delta y\;\;\;\;(14a)$

Taking the limits $\Delta x\rightarrow 0$ and $\Delta y\rightarrow 0$

$dz=\lim_{\Delta x\rightarrow 0}\left[\frac{f(x+\Delta x,y+\Delta y)-f(x,y+\Delta y)}{\Delta x}\right]\Delta x+\lim_{\Delta y\rightarrow 0}\left[\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\right]\Delta y$

Since the 1^st term on the RHS of the above equation is with respect to a change in $x$ , $y$ is a constant with $y+\Delta y\equiv y$ and

$dz=\lim_{\Delta x\rightarrow 0}\left[\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}\right]\Delta x+\lim_{\Delta y\rightarrow 0}\left[\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\right]\Delta y$

Substituting eq12 and eq13 in the above equation, we have eq14.

In general, the total differential of the function $z=f(x_1,x_2,\cdots,x_n)$ is

$dz=\left(\frac{\partial z}{\partial x_1}\right)_{x_2,x_3,\cdots,x_n}dx_1+\left(\frac{\partial z}{\partial x_2}\right)_{x_1,x_3,\cdots,x_n}dx_2+\cdots+\left(\frac{\partial z}{\partial x_n}\right)_{x_1,x_2,\cdots,x_{n-1}}dx_n$

If the variables themselves depend on another variable $t$ , i.e. $x=x(t)$ and $y=y(t)$ , we divide eq14a throughout by $\Delta t$ to give

$\frac{\Delta z}{\Delta t}=\left[\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x}\right]\frac{\Delta x}{\Delta t}+\left[\frac{f(x,y+\Delta y)-f(x,y)}{\Delta y}\right]\frac{\Delta y}{\Delta t}$

Since $\Delta x\rightarrow 0$ and $\Delta y\rightarrow 0$ as $\Delta t\rightarrow 0$ , if we take the limit $\Delta t\rightarrow 0$ , we have

$\frac{dz}{dt}=\left(\frac{\partial z}{\partial x}\right)_y \frac{dx}{dt}+\left(\frac{\partial z}{\partial y}\right)_x\frac{dy}{dt}\;\;\;\;\;\;\;\;(14b)$

Eq14b is known as the multivariable chain rule, which is also known as the total derivative of $z$ .

Next, we shall derive some of useful identities. With respect to eq14, if $y$ is a constant, $(dz)_y=\left(\frac{\partial z}{\partial x}\right)_y=(dx)_y$ , which when divided throughout by $(dz)_y$ gives $1=\left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial z}\right)_y$ or

$\left(\frac{\partial z}{\partial x}\right)_y=\frac{1}{\left(\frac{\partial x}{\partial z}\right)_y}\;\;\;\;\;\;\;\;(15)$

If $z$ is a constant, eq14 becomes $0=\left(\frac{\partial z}{\partial x}\right)_y(dx)_z+\left(\frac{\partial z}{\partial y}\right)_x(dy)_z$ , which when divided by $(dz)_z$ gives $0=\left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial y}\right)_z+\left(\frac{\partial z}{\partial y}\right)_x$ . Using the reciprocal identity of eq15, we have

$\left(\frac{\partial y}{\partial z}\right)_x\left(\frac{\partial z}{\partial x}\right)_y\left(\frac{\partial x}{\partial y}\right)_z=-1\;\;\;\;\;\;\;\;(16)$

If $dy=0$ in eq14b, we have the chain rule:

$\frac{dz}{dt}=\frac{dz}{dx}\frac{dx}{dt}\;\;\;\;\;\;\;\;(17)$

Finally, the second partial derivative of $z$ is defined as $\frac{\partial^2z}{\partial x\partial y}=\left[\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)_x\right]_y$ .

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February 7, 2023September 26, 2024

Irreversible work

Irreversible PV work falls under non-equilibrium thermodynamics, which is hard or sometimes impossible to calculate using simple equations. This is due to the difficulty of defining the properties of the system during the process. For example, during an irreversible expansion of a gas against a piston, the piston accelerates away from the system, resulting in regions of varying pressures in the system. To overcome this problem, broad assumptions are made to expresss irreversible PV work mathematically.

Consider the irreversible expansion of a system consisting of a gas in an isolated vertical cylinder. The frictionless piston, which is part of the surroundings, has mass $m$ and is held stationary with catches (see diagram above).

If the force exerted by the gas on the bottom surface of the piston is greater than the weight of the piston, the gas expands and pushes the piston up when the catches are removed. The piston moves over a distance $dx$ , until the force exerted by the expanded gas on the bottom surface of the piston equals to the weight of the piston. This implies that the gas is expanding against a constant force $F_{ext}=mg$ , which is due to the weight of the piston.

The change in energy of the surroundings is

$dU_{sur}=mgdx+dKE_{pist}\;\;\;\;\;\;\;\;(8)$

where $g$ is acceleration due to gravity and $KE_{pist}$ is the kinetic energy of the piston.

Since the initial and final $KE_{pist}$ are both zero, $dKE_{pist}=0$ and eq8 becomes

$dU_{sur}=mgdx\;\;\;\;\;\;\;\;(9)$

Noting that energy $U$ in the universe is conserved, where $dU_{sys}=-dU_{sur}$ , and that $mg=F_{ext}=p_{ext}A$ , eq9 becomes

$dU_{sys}=-p_{ext}dV$

Since the system is isolated, there is no transfer of heat. The change in energy of the system, according to the first law of thermodynamics, is therefore the change in work done on the system:

$dw_{irrev}=-p_{ext}dV\;\;\;\;\;\;\;\;(10)$

The integral form is:

$w_{irrev}=-\int_{V_i}^{V_f}p_{ext}dV=-p_{ext}\Delta V\;\;\;\;\;\;\;\;(11)$

If $m=0$ , the gas expands freely into the vacuum. In this case, $p_{ext}=0$ and therefore $w_{irrev}=0$ . In general, irreversible PV work against constant pressure is estimated using eq11.

Expansion work by a system on its surroundings is always greater when the process is carried out reversibly than irreversibly. This can be seen by plotting eq6 from the previous article and eq11 on the same PV graph (see upper diagram above), where the area under AC (reversible) is greater than the area under BC (irreversible). $p_{ext}$ for the irreversible process is made equal to the final pressure at $V_2$ when the piston stops, similar to our piston illustration above.

Conversely, compression work by the surroundings on the system is always greater when the process is carried out irreversibly (BA) than reversibly (CA, see lower diagram above). $p_{ext}$ for the irreversible process is now made equal to the final pressure at $V_1$ when the piston stops.

It is important to remember that the above is a crude attempt to associate an irreversible process with an equation.

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February 7, 2023September 26, 2024

Reversible isochoric process

A reversible isochoric process is a reversible thermodynamic process that occurs at constant volume. Consider an ideal gas in a piston-cylinder device immersed in a water bath. The piston is soldered to the cylinder walls and is immovable.

AD and BC in the above diagram are reversible isochoric processes. According to the ideal gas law, a system undergoing a reversible isochoric process from A to D requires the ratio of $\frac{T}{p}$ to be constant as the pressure of the system decreases. This is, in practice, carried out by continuously decreasing the temperature of the water bath by infinitesimal amounts. Since $dV=0$ , work done for a reversible isochoric process is zero (see eq5). Many chemical reactions that take place in a bomb calorimeter are carried out under isochoric conditions.

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Reversible isobaric process

A reversible isobaric process is a reversible thermodynamic process that occurs at constant pressure.

Specifically, a reversible isobaric process that follows the path of either A to B or D to C is a reversible isobaric expansion, while one that goes from either B to A or C to D is a reversible isobaric compression. We shall use the same setup of a piston-cylinder device immersed in a water bath as the one described in the previous article to illustrate reversible isobaric processes.

With reference to the ideal gas law, an isobaric expansion of a gas requires the ratio of $\frac{T}{V}$ to be constant as the volume of the system increases. Theoretically, for a reversible isobaric expansion to occur, the system has to be in contact with an infinite sequence of temperature baths, with one bath having a temperature that is infinitesimally higher than the next one. In practice, to achieve a result that is close to reversibility, the temperature of the water bath is continuously increased by infinitesimal amounts to maintain a temperature gradient across the cylinder walls so as to supply the system with energy for the expansion at constant pressure.

Since the pressure of the system remains constant and equals to that of the external pressure throughout the process, eq5 becomes

$w=-\int_{V_i}^{V_f}p_{ex}dV\;\;\;\;\;\;\;\;(7)$

For a reversible isobaric compression, the temperature of the water bath is continuously decreased by infinitesimal amounts to maintain a constant pressure in the system. Many chemical reactions that involve phase changes are conducted under isobaric conditions, e.g. $H_2O(s)\rightleftharpoons H_2O(l)$ .

Question

The change in work done is sometimes represented by $dw=d(pV)=pdV+vdp$ . How does this relate to eq6 and eq7?

Answer

The expression $dw=pdV+vdp$ or by convention $dw=-(pdV+vdp)$ is the general differential equation for work. If the change in pressure is infinitesimal, $dp\rightarrow 0$ , then the expression reduces to $dw=-pdV$ . If the process is a reversible isothermal process, we substitute the ideal gas law in $dw=-pdV$ to give $dw=-nRTln\frac{V_f}{V_i}$ , which is the differential form of eq6. If the process is isobaric, $dp=0$ , and $dw=-(pdV+vdp)$ again reduces to $dw=-pdV=p_{ex}dV$ , which is the differential form of eq7.

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February 7, 2023October 27, 2024

Reversible isothermal process

A reversible isothermal process is a reversible thermodynamic process that occurs at constant temperature.

A reversible isothermal expansion process for an ideal gas follows the path from A to C, while a reversible isothermal compression moves from C to A (see diagram above). The curve that describes an isothermal process is called an isotherm. We can express work done for a reversible isothermal process, which involves the change in pressure and volume of the system, by substituting the ideal gas law in eq5 to give:

$w=-nRT\int_{V_i}^{V_f}\frac{1}{V}dV=-nRTln\frac{V_f}{V_i}\;\;\;\;\;\;\;\;(6)$

To determine the work done on a system undergoing reversible isothermal expansion or compression, we immerse the piston-cylinder device in a constant temperature water bath. For a reversible isothermal compression, the external force balances the force exerted by the gas on the underside of the piston before the system is compressed. At the start of the compression, the external force is increased infinitesimally causing the piston to move down over an infinitesimal distance.

Consequently, infinitesimal energy is transferred via work to the system, leading to an infinitesimal increase in the kinetic energy of the gas and hence, an increase in the system’s temperature. The infinitesimal difference in temperature between the system and the water bath results in the flow of energy from the system through the thermally conducting cylinder walls to the temperature controlled water bath. Hence, constant temperature of the system is maintained throughout the process. This compression process is repeated stepwise with further infinitesimal increases in the external force until the desired final volume of the system is achieved. Work done is then calculated using eq6. If the external force is now decreased infinitesimally in a stepwise manner, we have a reversible isothermal expansion of the gas. Many chemical reactions involving the determination of rates of reactions are carried out isothermally.

Question

Does the temperature of the water bath increase if energy is continuously transferred from the system to the water bath in the compression process?

Answer

Theoretically, the constant temperature water bath is infinitely large. This means that transfer of energy from the system to the layer of water in the vicinity of the system (due to a temperature difference between the two) is the first of infinite steps of transfer, with the second step being the transfer of energy from the body of water in the vicinity of the system to a layer of water further away from the system and so on. In practice, the temperature of the water bath is controlled by a computer and adjusted throughout the process to maintain constant temperature for the system.

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