WORK
Work (w) is not a state function. The amount of work done from a fixed amount of heat is not a unique value. It depends on the process used. For the same amount of heat energy, a more efficient process will produce more work.
Since work is not a state function, heat (q) cannot be a state function according to the First Law of Thermodynamics, ΔU = q + w. That is q is dependent on the value of w, since U is independent of the process.
Mechanical work in physics is defined as energy needed to move against a force. If the system moves a distance of x against an opposing force of F, then the amount of work done by the system is;
Note the negative sign for work done by the system.
Tutorial 1
Compute the work done when 55.85g (or 1 mole) of iron reacts with excess hydrochloric acid in a piston reactor (that is the lid can move like a piston) at SATP. Assume the temperature remains constant. Answer
HEAT
Heat given to a system need not necessary produces work, it can add to the internal energy of the system. Let us consider tutorial 1 above. Instead of a piston reactor we use an immoveable reactor of fixed volume. Then even when the hydrogen gas is given off no mechanical work is being done. Logic says we can define another term H, known as enthalpy, so that;
This section of the lesson is like one of those movies when you start from the end and work up to the beginning. At the end you will appreciate why H is defined as such.
U, p and V are all state functions, so H must also be a state function. A change in H will be according to;
| ∆H | = | ∆U + ∆(pV) | |
| = | ∆U + p∆V + V∆p | ||
| Or | ∆U | = | ∆H − p∆V ; for processes where p is kept constant |
This is the First Law of Thermodynamics with q = ∆H. So by expressing enthalpy the way we did, we have found a condition for q to be a state function. This condition is no great deal as most of the chemistry done in the laboratory is at one pressure; atmospheric pressure.
Like energy and work we do not know exactly what enthalpy is. We can only rewrite the expression into words hoping it says something. When the process on a system is carried out at constant pressure and allowed to do mechanical work, then the heat energy given to the system is used to change the enthalpy. It does not tell us anything new, but it does make us think when you try to say it. Even though this is the case you will eventually appreciate how much the concept of enthalpy contributed to applied chemistry.
Tutorial 2
When an electric kettle is used to boil water it was found that 40.6 kJ of electrical energy was needed to vapourise 18g of water. Compute the change in enthalpy and in the molar internal energy at the boiling point. Answer
HEAT CAPACITY
When you read thermodynamics in most chemistry textbooks you will be hypnotised into thinking that it is all about gases. Even that it focus on ideal gas. Like I say it is hypnotism. Thermodynamics is about understanding the universe. However ideal gas seems an easy way to explain most of the concepts, otherwise the subject would appear very abstract. It is assumed that there is no internal interaction in ideal gas. If we used liquids or solid to explain mechanical work we have to go into the thermal expansion coefficient, isothermal compressibility, etc. This is of course not so familiar as the perfect gas equation.
Having said that you will notice that I am attempting to present thermodynamics for all matters; solid, liquid and gas. Because of this I have skipped many topics involving gases. If you have problem with them please write me.
One parameter to measure how heat energy affect the internal energy of matter is the molar heat capacity; the amount of heat needed to increase the temperature of one mole of matter by one degree. (Chemists like to think in terms of atoms and molecules and so we use molar heat capacity, other field of science and engineering generally use per unit mass of matter instead of mole and this is known as specific heat capacity.) The mathematical expression would be; C = q / (n x ∆T)
Again we must be aware that the heat involved will not be the same at constant volume and constant pressure. Let us refer to the First Law of Thermodynamics; U = q + p∆V. At constant pressure q = U - p∆V, while at constant volume it will be q = U. So we have to differentiate the two.
We define molar heat capacity at constant volume as; Cv = {∂U / ∂T}v,m
And molar heat capacity at constant pressure as; Cp = {∂H / ∂T}p,m
The next question to ask will be; is the value of the molar heat capacity for a matter the same at all temperatures? That is the amount of heat one mole of the matter can absorb when you heat it from 100�C to 101�C (for example) be the same as when you heat it from 500�C to 501�C (for example). It is very obvious that it should not be. However the variation of the molar heat capacity with temperature was not great and it is not convenient to determine the value for each and every temperature. It is good enough to use a series expression to correct for the slight variation. For example Cp,m = a + bT + cT‾� and determine value of the constants a, b, and c for the matter.
U is a function of (P,V,T), and (P,V,T) are interrelated. Let us consider U as a function of (V,T) (this will automatic include T) then according to mathematics;
| dU | = | ∂U (------)TdV ∂V |
+ | ∂U (------)vdT ∂T |
U is energy and pdV is also energy, so dU/dV is equivalent to pressure which we will represent by π
| dU | = | πTdV | + | CVdT |
| Or | ∂U (-----)p ∂T |
= | ∂V πT(-----)p ∂T |
+ | CV |
Tutorial 3 What is the value for πT of an ideal gas. Answer
When a matter is heated it expands, the expansion coefficient (α) is defined as;
| α | = | 1 ∂V ---- (-----)p V ∂T | |
| So |
∂U (-----)p ∂T |
= | απTV + CV |
There is no problem determining Cp,m in a laboratory, but Cv,m is a real problem as it involves change in internal energy. This section will try to find a way to do this.
| H | = | U + pV |
|
∂H (-----)p ∂T |
= | ∂U (-----)p + ∂T |
∂(pV) (---------)p ∂T | = | απTV + | CV + | ∂(pV) (---------)p ∂T |
| Cp - Cv | = | απTV + | ∂(pV) (---------)p ∂T |
Tutorial 4 Compute Cp,m + Cv,m for an ideal gas. Answer
| ∂(pV) (---------)p ∂T |
= | ∂V p(----)p = αpV ∂T | |
| So | Cp − Cv | = | α(p + πT)V |
| We will show much, much later that; | πT | = | ∂p T(-----)V − p ∂T |
For partial derivatives Euler's chain relation states; |
∂x ∂y ∂z (-----)z(-----)x(-----)y ∂y ∂z ∂x |
= | − 1 |
| So | ∂p ∂T ∂V (-----)V(-----)p(-----)T ∂T ∂V ∂p |
= | − 1 |
| Defining isothermal compressibility of matter as; | kT | = | 1 ∂V − ---- (-----)T V ∂p |
| ∂p (− kT) (-----)V ------ ∂T α |
= | − 1 | |
| And | Cp − Cv | = | (α� TV) / kT |
Tutorial 5
Given that at SATP the density of the liquid tetrachloromethane (CCl4) is 1.59 g cm‾�, the thermal expansion coefficient 1.24 x 10‾� K‾�, the isothermal compressibility 9.05 x 10‾5 atm‾�, and the molar heat capacity at constant pressure 132 JK‾� mol‾�, compute the change in the internal energy of tetrachloromethane for each degree increase in temperature at constant volume? Answer
CALORIMETRY
 |
| T = Thermometer IS = Ignition System U-D S = Up-Down Stirrer W = water S = Sample AE = Adiabatic Enclosure BR = Bomb Reactor |
The reaction is conducted in the reactor. The heat energy given off increases the temperature of the water, and is measured. Of course the surrounding in this case is the water and the insulated enclosure.
We always begin the experiment by calibrating the calorimeter. This is done by using a standard reaction with known heat of reaction. Then using the calorimeter with the amount of water needed for the actual experiment, we carry out the reaction. Since the calorimeter is thermally insulated all the heat energy must stay in the calorimeter, increasing the temperature of whatever is inside. A thermometer is placed in the water and protruding out of the calorimeter for you to monitor the temperature change. Normally the calorimeter comes equipped with some sort of stirring mechanism for the water to help it achieve thermal equilibrium faster with very gentle stirring. Once the temperature comes to a fairly constant reading, we tell ourselves that the whole calorimeter has achieved thermal equilibrium, and so the increase in temperature represents the known amount of heat given off by the reaction.
Now we conduct the actual experiment. With the increase in temperature determined we work backward to compute the amount of heat given off by the reaction we are studying. So you can see why it is important to use the same amount of water for the calibration as in the experiment.
