- +1-530-264-8006
- info@tutorsglobe.com

18,76,764

Questions

Asked

21,311

Experts

9,67,568

Questions

Answered

Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!

Submit Assignment
## Types of processes, Chemistry tutorial

:IntroductionWhenever the state of a system changes, it is stated to have undergone a process. Therefore a process signifies change in at least one of the state variables of the system. The procedure might be accompanied via an exchange of matter and energy between the system and the surroundings. There are some methods in which a specific state variable (that is, thermodynamic property of the system) remain unchanged. These processes are of the given types:

Isothermal process:

In an isothermal process, the temperature of system remains constant. Whenever a system experiences an isothermal process, it is in thermal contact by a large constant temperature bath, termed as thermostat. The system maintains its temperature through exchange of heat by the thermostat.

Adiabatic process:

In the adiabatic process, no heat is allowed to enter or leave the system. The systems in which such processes take place are thermally insulated from the surroundings. An adiabatic process might involve increase or decrease in the temperature of system.

Isobaric process:

The isobaric process is one in which the pressure of the system remains unaffected or unchanged. A reaction occurring in an open beaker is always at atmospheric pressure and therefore, the process is isobaric.

Isochoric process:

In an isochoric process, the volume of system remains constant. Therefore, a chemical of constant volume is an isochoric process.

Cyclic process:

A process is stated to be cyclic if the system (after any number of intermediate changes) returns to its original state. The initial and final value of each and every thermodynamic variable is similar after the completion of a cyclic process. Based on the value of the driving force applied we can categorize the processes into two kinds, namely, reversible and irreversible.

a) Reversible process: A reversible process is one in which at any time, the driving force only surpasses the opposing force very slightly, and therefore, the direction of the process can be reversed by just a small change in a variable such as temperature and pressure. The idea of a reversible procedure will become clear by considering the giving example.

Let's take a gas at pressure 'p' in a cylinder fitted by an air-tight piston. If the external pressure on the gas is equivalent to the pressure of the gas, then there is neither expansion nor compression and the piston remains at its position. Though, on increasing the external pressure (p

_{ext}) infinitesimally, the gas can be compressed. On the other hand by slightly reducing the external pressure, the gas might be expanded. Therefore,Note that 'if' is common to all the three statements.

If p

_{ext}= p, then the system is static and piston doesn't move.If pext = p + dp, then the gas is compressed and the piston moves downwards infinitesimally slowly.

If P

_{ext }= p - dp, then the gas expands and the piston moves outwards infinitesimally slowly.Therefore, we can notice that in a reversible process, the direction of a process is changed via changing the magnitude of the driving force via a small amount.

b) Irreversible process: Any procedure which is not reversible is known as irreversible. All the natural processes are irreversible. The flow of heat from a high temperature body to a low temperature body is a natural process and thus, irreversible. Therefore, the expansion of a gas against the vacuum, termed as free expansion. Irreversible processes are as well termed as spontaneous processes.

:Work, Heat and Heat capacityWork, heat and energy encompass the similar units, namely Joule (J). Energy is the thermodynamic property of a system, while work and heat are not. The latter two are meaningful only whenever a process occurs. Let us first define heat and work.

Heat: Heat is the form of energy. Heat is not the property of a system however is exchanged between a system and the surroundings throughout a process, whenever there is a temperature difference between the two.

Work: Work (W) is stated as the product of the force applied (F) and the distance (X) moved all along the direction of the force.

W = F.X

Forces have various physical origins and work can be completed or done in a variety of ways:

A) Gravitational Work: Whenever a body of mass 'm' is moved through a height 'h' and gravity 'g', then force is equivalent to mg and the gravitational work done is mgh.

B) Electrical work: If an electric potential 'E' is applied across a resistance 'R' in such a way that current 'i' flows through it, then work done per second is 'Ei' and in t seconds it is equivalent to 'Eit'.

C) Pressure-volume work: This is a kind of mechanical work performed whenever a system changes its volume against the opposing pressure. This as well is expansion or compression.

The energy lost or gained during heat exchange between the system and the surroundings can be defined in terms of heat capacity values.

Define Heat capacity:

Heat capacity is the heat needed to increase the temperature of a body by 1K. If, throughout the process, the volume of the system remains constant, then it is known as heat capacity at constant volume (C

_{v}); if the pressure remains unchanged, it is known as the heat capacity at constant pressure (C_{p}). For one mole of a pure substance, these are termed as molar heat capacity at constant pressure, C_{p}and at constant volume C_{v}. The heat capacities per unit mass are known as specific heats. The heat capacities change by temperature. This signifies that, the heat needed to change the temperature by 1 K is different at various temperatures. Though, over small ranges of temperature, these are generally taken as constant. The molar heat capacity and specific heat are intensive properties whereas the heat capacity is an extensive property.For changing the temperature of a specific system by dT, if the heat needed is dqv (at constant volume) or dqp (at constant pressure), then we encompass

C

_{V}= nC‾_{V}= dq_{V}/dTC

_{P}= nC‾_{P}= dq_{P}/dTHere, 'n' is the amount (that is, number of moles) of the substance comprising the system.

From the above equations, it is possible to find out the heat needed for a process, by integration over the temperature range T

_{1 }and T_{2}. Therefore,q

_{V}=_{T1}∫^{T2}C_{V}dT =_{T1}∫^{T2}nC‾_{V }dTqV =

_{T1}∫^{T2}C_{q}dT =_{T1}∫^{T2}nC‾_{V}dTFormulas to solve numerical problems:

Formula 1: If m is not is equivalent to -1

_{x1}∫^{x2}ax^{m}dx = a_{x1}∫^{x2 }x^{m}dx = a/(m + 1)[x^{m+1}]_{x1}^{x2}= a/(m + 1)[x_{2}^{m+1 }- x_{1}^{m+1}]Here, 'a' is a constant.

Formula 2: If 'm' is equivalent to -1

_{x1}∫^{x2 }a (dx/x) = a ln (x_{2}/x_{1})Again, 'a' is a constant. Note that 'ln' signifies logarithm to the base e. As we use natural logarithm (that is, logarithm to the base 10) in our computations, it is better to modify formula 2 as:

_{x1}∫^{x2}a (dx/x) = 2.303 a log (x_{2}/x_{1})It will be noted that ln x = 2.303 log x

:The First law of ThermodynamicsThe first law of thermodynamics was first introduced by Mayer and Helmholtz in the year 1840 in Germany, Joule in England, and Colding in Denmark. This law is as well termed as the law of conservation of energy.

The first law of thermodynamics can be defined in any one of the given ways:

a) Energy of the isolated system remains constant.

b) Energy can neither be created nor destroyed however it can be changed from one form to the other.

c) This is not possible to form a perpetual motion machine that can work endlessly without the expenses of energy. (Such a machine is termed as perpetual motion machine of the first type.)

All the above statements are equal to the principle of conservation of energy. These statements indicate that energy of a system will remain constant if it is left undisturbed. If, on the other hand, the system interacts by the surroundings, then its energy might change; however then, there will as well be equivalent and opposite change in the energy of the surroundings. As Work is a form of energy, it is not possible for a machine to keep on doing work for ever. As soon as its own energy is exhausted it will need a source of energy to carry on doing work. The first law of thermodynamics; has no theoretical evidence. This is a law which is completely based on observation. As the law has never been contradicted, its truth is taken for granted.

Einstein in the year 1905 illustrated that matter and energy are interconvertible according to the equation,

E = mc

^{2}This signifies that if mass 'm' is destroyed, then energy 'E' is produced; 'c' is the velocity of light. This is not a contradiction as mass can as well be considered as a form of energy.

Internal Energy:

The internal energy 'U' of a system is the total energy of its constituent atoms or molecules comprising all forms of kinetic energy (that is, due to translation, vibration and rotation) and also due to all kinds of interactions between the molecules and sub-particles. This is a state variable and an extensive property and its absolute value can't be found out. Though, in thermodynamics we are interested, in the changes in internal energy that accompany any process, however not in the absolute value of 'U'.

Such changes in the internal energy can be brought about only through interaction of the system by its surroundings. The change in internal energy (ΔU) based only on the initial and final states and is Independent of the path adopted.

The net energy of a system is the sum of the internal energy and some energy due to the motion or position of the system as a whole. For illustration, the net energy of water on the ground floor is different from its energy on the top of the building. The difference in total energy is mgh (here, 'm' is the mass of water, 'h' the height of the building and 'g' the acceleration due gravity) while the internal energy is similar; in both cases. Likewise for a moving bullet, the net energy is the sum of the internal energy and its kinetic energy, 1/2 mv

^{2 }(here, 'm' is the mass of the bullet and 'v' its velocity).The heat change, dq, and the work done, dw, contribute towards the change in the internal energy of the system. We adopt the convention that the change in the internal energy is equivalent to the sum of the heat absorbed via the system and the work done on it.

If dq is positive (that is, heat absorbed by the system is positive), this leads to an increase in the internal energy of the system. A negative dq involves loss of heat from the system and % signifies a decrease in the internal energy.

If the work done on the system (dw) is positive, this rises the internal energy of the system.

A negative value of dw entails that work has been completed by the system at the expense of its internal energy.

Let us now derive the arithmetical forms of the first law of thermodynamics.

=> Mathematical Forms of the First Law of Thermodynamics:

As illustrated earlier, work and heat bring about changes in the internal energy of the system. Whenever the system absorbs a certain amount of heat, dq then its internal energy increases by this amount; moreover if dw is work done on the system, then the internal energy again increases. Therefore, the total change in the internal energy is represented by:

dU = dq + dw

The above equation is for infinitesimal changes. If, though, a system is taken from its initial state to the other state via a process in which the heat absorbed is q and work done on the system is w, then the total change in internal energy will be represented by:

ΔU = U

_{f }- U_{i }= q +Here, U

_{i}and U_{f}are the initial and final internal energies of the system and ΔU, the total change. It is clear that ΔU can encompass either negative or positive value based on q and w. However once the initial and final states of a system are fixed, ΔU is automatically fixed, no matter what path is adopted in carrying out the procedure. In another words, internal energy is a state function whereas heat and work are not. As an illustration consider a system being taken from the initial state where it has energy U_{i}to the final state having energy U_{f}all along the different paths I, II and III (figure shown below). Then in all such cases, ΔU is similar. If this were not so, then it would have been possible to make a perpetual motion machine by reaching the final state via a high energy change path (state, III) and coming back through a low-energy change path (IV) thus releasing the energy difference for work. The above two equations are arithmetical expressions of the first law of thermodynamics.Fig: First law of thermodynamics paths

:Isothermal expansionIn this part, we are going to compute the work done on the gas in an isothermal process. For this we should first arrive at a general expression for pressure - volume work done in an air tight piston of area 'A'. Suppose that pressure of the gas is 'p' and the external pressure is 'p

_{ext}' that is slightly less than the gas pressure. The gas will, therefore, expand against the opposing force which is represented by:F = P

_{ext}AIf for the period of expansion, the piston moves via a distance dx, then this small amount of work dw' done by the system is represented by:

dw' = fdx = P

_{ext }AdxHowever, Adx is the change in volume of the gas, dV. Therefore,

dw' = P

_{ext}dVTherefore, the work done by the system (that is, by the gas) is dw'. Thus, the work done on the gas is - dw' which we shall represent by dw. Therefore, work dw done on the system is:

dw' = - p

_{ext}dVThe above equation is a general expression helpful in computing pressure - volume work whether it is isothermal or adiabatic process. This can be observed that dw is negative if the gas expands and it is positive if the gas contracts, [dv = +ve in expansion and dv = -ve in compression].

We shall now compute the work of expansion (and as well of compression) under isothermal conditions. Let first take up the work done under isothermal irreversible conditions.

Isothermal Irreversible Process:

Let us suppose that the gas kept in a cylinder expands isothermally and irreversibly against the constant process.

This signifies that:

a) Gas expands against the constant External pressure (p

_{ext }= constant)b) There is a considerable difference between the gas pressure (within the cylinder) and the external pressure.

c) That the temperature doesn't change throughout the process.

Assume that the initial and final volumes be V

_{1}and V_{2}, correspondingly. The total work, 'W', done on the system is obtained by integrating the equation dw' = - p_{ext}dV.W = -

_{V1}∫^{V2}P_{ext }dV = - P_{ext}_{V1}∫^{V2 }dV= - P

_{ext }(V_{2}- V_{1}) = - P_{ext}ΔVThe symbol, ΔV, represents the total change in volume throughout the process.

Isothermal Reversible Process:

We are already familiar that a reversible process can be taken out whenever external pressure (p

_{ext}) is only infinitesimally different from the gas pressure within a cylinder (p)In such a case, p

_{ext}≈ p and therefore, dw' = - p_{ext}dV can be written as:dw = - pdV

The total work done, 'W', as the gas expands isothermally and reversibly from V

_{1}to a volume V_{2}is then given by integrating dw = - pdV within limits V_{1}and V_{2}.W = -

_{V1}∫^{V2 }pdVLet us suppose that the gas behaves ideally. Therefore,

p = nRT/V

Using the above equation that is, p = nRT/V on W = -

_{V1}∫^{V2 }pdV, we get:W = -

_{V1}∫^{V2}(nRT/V) dV= nRT -

_{V1}∫^{V2}(dV/V)= - nRT ln (V

_{2}/V_{1}) = nRT ln (V_{1}/V_{2})Therefore, W = - 2.303 nRT log (V

_{2}/V_{1}) = 2.303 nRT log (V_{1}/V_{2})It can be observed that if V

_{2}is less than V_{1}, then the gas has been compressed and, 'W' is positive. As well, the value of 'W' then happens to be the minimum work needed for compressing the gas from (volume) V_{1}to V_{2}.Likewise, if V

_{2}is greater than V_{1}then the gas experiences expansion and 'W' is negative.This signifies, work is done by the gas - W symbolizes the maximum work available via expansion.

The equation W = - 2.303 nRT log (V

_{2}/V_{1}) = 2.303 nRT log (V_{1}/V_{2}) can also be given in terms of initial and final pressures (p_{1 }and p_{2}) of the ideal gas.For an ideal gas at constant temperature,

p

_{1}V_{1 }= p_{2}V_{2}V

_{2}/V_{1}= p_{1}/p_{2}W = - 2.303 nRT log (p

_{1}/p_{2})= 2.303 nRT log (p

_{2}/p_{1})Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online Chemistry tutoring. Chat with us or submit request at info@tutorsglobe.com