The first process is an isothermal expansion , with the volume of the gas changing its volume from. This isothermal process is represented by the curve between points A and C. The gas is kept at a constant temperature T by keeping it in thermal equilibrium with a heat reservoir at that temperature. From Figure and the ideal gas law,. The expansion is isothermal, so T remains constant over the entire process.
Since n and R are also constant, the only variable in the integrand is V , so the work done by an ideal gas in an isothermal process is. Notice that if expansion , W is positive, as expected. The straight lines from A to B and then from B to C represent a different process. Here, a gas at a pressure first expands isobarically constant pressure and quasi-statically from , after which it cools quasi-statically at the constant volume until its pressure drops to.
From A to B , the pressure is constant at p , so the work over this part of the path is. From B to C , there is no change in volume and therefore no work is done. The net work over the path ABC is then. A comparison of the expressions for the work done by the gas in the two processes of Figure shows that they are quite different.
This illustrates a very important property of thermodynamic work: It is path dependent. We cannot determine the work done by a system as it goes from one equilibrium state to another unless we know its thermodynamic path. Different values of the work are associated with different paths. Isothermal Expansion of a van der Waals Gas Studies of a van der Waals gas require an adjustment to the ideal gas law that takes into consideration that gas molecules have a definite volume see The Kinetic Theory of Gases.
One mole of a van der Waals gas has an equation of state. Suppose the gas expands isothermally and quasi-statically from volume to volume How much work is done by the gas during the expansion? Strategy Because the equation of state is given, we can use Figure to express the pressure in terms of V and T.
Furthermore, temperature T is a constant under the isothermal condition, so V becomes the only changing variable under the integral.
Solution To evaluate this integral, we must express p as a function of V. From the given equation of state, the gas pressure is. Because T is constant under the isothermal condition, the work done by 1 mol of a van der Waals gas in expanding from a volume to a volume is thus.
Significance By taking into account the volume of molecules, the expression for work is much more complex. If, however, we set and we see that the expression for work matches exactly the work done by an isothermal process for one mole of an ideal gas.
The internal energy of a thermodynamic system is, by definition, the sum of the mechanical energies of all the molecules or entities in the system. If the kinetic and potential energies of molecule i are and respectively, then the internal energy of the system is the average of the total mechanical energy of all the entities:.
The potential energy is associated only with the interactions between molecule i and the other molecules of the system. GOB Video Lessons. Microbiology Video Lessons. Calculus Video Lessons. Statistics Video Lessons. Microeconomics Video Lessons. Macroeconomics Video Lessons. Accounting Video Lessons. This energy is measured by burning food in a calorimeter, which is how the units are determined.
Catabolism is the pathway that breaks down molecules into smaller units and produces energy. Anabolism is the building up of molecules from smaller units. Anabolism uses up the energy produced by the catabolic break down of your food to create molecules more useful to your body. Our body loses internal energy, and there are three places this internal energy can go—to heat transfer, to doing work, and to stored fat a tiny fraction also goes to cell repair and growth.
As shown in Fig 1 heat transfer and doing work take internal energy out of the body, and then food puts it back. If you eat just the right amount of food, then your average internal energy remains constant.
The reverse is true if you eat too little. This process is how dieting produces weight loss. Metabolism : a The first law of thermodynamics applied to metabolism. Heat transferred out of the body Q and work done by the body W remove internal energy, while food intake replaces it.
Food intake may be considered as work done on the body. Life is not always this simple, as any dieter knows. The body stores fat or metabolizes it only if energy intake changes for a period of several days.
Once you have been on a major diet, the next one is less successful because your body alters the way it responds to low energy intake. Your basal metabolic rate is the rate at which food is converted into heat transfer and work done while the body is at complete rest. The body adjusts its basal metabolic rate to compensate partially for over-eating or under-eating. The body will decrease the metabolic rate rather than eliminate its own fat to replace lost food intake.
You will become more easily chilled and feel less energetic as a result of the lower metabolic rate, and you will not lose weight as fast as before. Exercise helps with weight loss because it produces both heat transfer from your body and work, and raises your metabolic rate even when you are at rest.
The body provides us with an excellent indication that many thermodynamic processes are irreversible. An irreversible process can go in one direction but not the reverse, under a given set of conditions. For example, although body fat can be converted to do work and produce heat transfer, work done on the body and heat transfer into it cannot be converted to body fat. Otherwise, we could skip lunch by sunning ourselves or by walking down stairs. Another example of an irreversible thermodynamic process is photosynthesis.
This process is the intake of one form of energy—light—by plants and its conversion to chemical potential energy. Both applications of the first law of thermodynamics are illustrated in. One great advantage of such conservation laws is that they accurately describe the beginning and ending points of complex processes such as metabolism and photosynthesis without regard to the complications in between.
Privacy Policy. In the course of the reaction, heat is either given off or absorbed by the system. Furthermore, the system either does work on it surroundings or has work done on it by its surroundings. Either of these interactions can affect the internal energy of the system. Two kinds of work are normally associated with a chemical reaction: electrical work and work of expansion.
Chemical reactions can do work on their surroundings by driving an electric current through an external wire. Reactions also do work on their surroundings when the volume of the system expands during the course of the reaction The amount of work of expansion done by the reaction is equal to the product of the pressure against which the system expands times the change in the volume of the system.
The sign convention for this equation reflects the fact that the internal energy of the system decreases when the system does work on its surroundings. What would happen if we created a set of conditions under which no work is done by the system on its surroundings, or vice versa, during a chemical reaction? Under these conditions, the heat given off or absorbed by the reaction would be equal to the change in the internal energy of the system.
The easiest way to achieve these conditions is to run the reaction at constant volume, where no work of expansion is possible. At constant volume, the heat given off or absorbed by the reaction is equal to the change in the internal energy that occurs during the reaction. The figure below shows a calorimeter in which reactions can be run at constant volume. Most reactions, however, are run in open flasks and beakers. When this is done, the volume of the system is not constant because gas can either enter or leave the container during the reaction.
The system is at constant pressure, however, because the total pressure inside the container is always equal to atmospheric pressure. If a gas is driven out of the flask during the reaction, the system does work on its surroundings. If the reaction pulls a gas into the flask, the surroundings do work on the system. We can still measure the amount of heat given off or absorbed during the reaction, but it is no longer equal to the change in the internal energy of the system, because some of the heat has been converted into work.
We can get around this problem by introducing the concept of enthalpy H , which is the sum of the internal energy of the system plus the product of the pressure of the gas in the system times the volume of the system. For the sake of simplicity, the subscript "sys" will be left off the symbol for both the internal energy of the system and the enthalpy of the system from now on.
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