Arrange the entropy transforms of the adhering to processes, every at $25^circ mathrmC$, in the intended order of enhancing $Delta S,$ and also explain her reasoning:(a) $mathrmH_2 mathrmO(mathrml, 1 mathrmbar) longrightarrow mathrmH_2 mathrmO(mathrmg, 1 mathrmbar)$(b) $mathrmCO_2(mathrms, 1 mathrmbar) longrightarrow mathrmCO_2(mathrmg, 0.01 mathrmbar)$(c) $mathrmH_2 mathrmO(1,1 mathrmbar) longrightarrow mathrmH_2 mathrmO(mathrmg, 0.01 mathrmbar)$
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we must rank these three reactions in bespeak of increasing entropy. First, you deserve to make some assumptions around the system. We have the right to say the it's a closed system, so our temperature will remain constant. And also we can assume that all the gases monitor the best gas legislation PV equates to in our tea. The very first we deserve to compare A and also C. They have actually very comparable reactions. We have water starting as a liquid and also at one bar goes come water as a gas. And also this is where it's various is part a stays at a continuous pressure. In part, C reduce in pressure. So just how does a readjust in pressure impact the entropy of the system? First, let's look at the right gas law. You know, our sister Constant. Due to the fact that we it seems ~ to continue to be closed system, we deserve to say that the temperature continues to be constant. Instance, the reaction is a 1 to 1 molar ratio. We have the right to say the the variety of moles stays consistent as well, so we have PV amounts to some constant. So this tells united state that as push decreases, volume will have to increase, and we know that when volume increases, the molecules have an ext space to relocate around and also this method that the readjust in entropy is positive. So in component C, us do have a diminish in pressure. Well, part remains constant. So you understand that part C will have a greater change in entropy 보다 party. We're spring at part B. We have the right to compare this to component C. We have carbon dioxide in this. A solid in ~ one bar goes come carbon dioxide as a gas in ~ 10.1 bar. Therefore the only distinction between part B and also C is the step change. We know that a solid has the least mobility, the least entropy of all the deals with of matter. And also going to a guess: v is the most entropy of every the faces of matter.
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For this reason the adjust in entropy from a solid to a guess: v is higher than the interview adjust from a liquid to a gas. So since they both starting finish in ~ the very same pressure, we can say that the entropy change for B is greater than see