R ⋅ T ⋅ ln 2, while work of the mixing of the identical gases was equal to zero. In this article, the work of the isothermal mixing of the two different gases, taken in quantity of one mole each, was determined to be equal to ![]() As far as we know, according to, the first attempt to use half-penetrable membranes for mixing two different gases was made by Wiedeburg. Having marked the impossibility of the reversible process at an infinite difference in the gravitational potentials of different gases, we shall now consider solutions of the Gibbs’ paradox, which are based on separating different gases with half-penetrable membranes. We should note here that an infinite difference of the gravitational potential could not be used because it contradicts the general relativity theory. ![]() Also, modern understanding of nature prohibits such a process. To tell the truth, we have never found any valid suggestion for this process using an infinitely strong gravitational field for separating different gases in a limited long tube. That is why the process of using infinite shifts of the substance could not be realized, either. Actually, a limited speed of any shift of the substance is an experimentally proven fact. However, we cannot accept a hypothesis allowing a real process with infinite shifts. According to his calculation it appeared that the mixing entropy was constant for different gases and equal to zero for identical gases. He suggested using an infinitely long tube situated in the gravitational field to separate out the different gases from a mixture. If we have to make an infinite number of the infinitely long processes then this would not provide a reversible condition of this process at all. This theory would actually require that, each reversible process be infinitely long. We cannot agree with the possibility that this is a reversible process, where an infinite number of cycles were used. ![]() R ⋅ ln 2 was obtained for mixing entropy of different gases taken in the quantity of 1/2 mole each. Certainly, in the result of this process, the value of It was suggested that an infinite number of cycles would have to be used to separate the pure gases from a mixture. Using this technique, these gases could not be completely separated because of the limited height of the apparatus. Rayleigh suggested using the gravitational field for separating two different gases from a mixture. This difference is small, at 0.74 J/K.Similarly, the first attempt to find the work of mixing two different gases belongs to the Lord Rayleigh. The change in entropy ∆S is the difference between the entropy lost by the lake and that gained by the ice. You don’t use ln terms because the temperature is constant. Here the entropy of the lake uses a different tactic. Now what happens to the entropy of the lake? The lake has constant temperature of 288K. Then the final section is where the temperature goes from 0 to 10 C. There is the part where the temperature goes from -10 to 0 C, then there is the heat of fusion of the water section. Notice there are three parts to the above problem. Then we use the method of cutting into pieces we talked about earlier. ![]() We first assume the final temperature remains constant at 15 C. This is a problem out of the Serway Physics book problems section. Calculate the change in entropy of the cube-lake system as the ice cube comes to thermal equilibrium with the lake. Trying to add steps, like taking the temperature difference of an entire process of several steps and dividing the average heat value of the whole process by this temperature change won’t accomplish anything.įor our first example we have a 10 g ice cube at -10 C that is put into a lake whose temperature is 15 C. This cut and paste method is the only way to get the correct answer. Basically you cut up each part of a phase diagram into a certain section and solve for the entropy of each of the sections by themselves and then add them together at the end. Generally you use the ln formula when the temperature changes and you use mL for fusion changes. If the temperature doesn’t change you make T constant, but the heat could change under certain circumstances like described earlier. When do you use mCvln(Tf-Ti) to find the heat? You use this if the temperature changes.
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