How to find out max useful work in thermo dynamics?
Answers
Answered by
1
HOLA MATE here is u r answer....
Negative of the ΔGΔG for a process is the maximum useful work that can be obtained from it (at constant pressure and temperature). I understood this in this way: ΔHΔH is the heat absorbed by the system (since the process is at constant temperature and pressure), so equivalently −ΔH−ΔHenergy is obtained from the system after doing expansion work. Since ΔSΔS is the entropy created in the process, at the very least −ΔS−ΔS entropy must be created in the surroundings - that is, at least −TΔS−TΔS energy must be lost as heat. This comes from the −ΔH−ΔH, and thus leaves −ΔH+TΔS−ΔH+TΔS to do useful work. No more work can be done than this. −ΔH+TΔS−ΔH+TΔS is −ΔG−ΔG, so −ΔG−ΔG is the maximum possible useful work. First of all, I wanted to know if this is correct, and if this is actually why −ΔG−ΔG is the maximum possible useful work.
Now if |ΔH||ΔH| is more than |TΔS||TΔS|, with both being negative, then one can think of the WmaxWmax or −ΔG−ΔG as a part of the |ΔH||ΔH|; since |ΔG|<|ΔH||ΔG|<|ΔH|. Some part of |ΔH||ΔH| goes as |TΔS||TΔS| to increase the surroundings' entropy, and the other part in doing useful work.
Hope it helps uuu
Negative of the ΔGΔG for a process is the maximum useful work that can be obtained from it (at constant pressure and temperature). I understood this in this way: ΔHΔH is the heat absorbed by the system (since the process is at constant temperature and pressure), so equivalently −ΔH−ΔHenergy is obtained from the system after doing expansion work. Since ΔSΔS is the entropy created in the process, at the very least −ΔS−ΔS entropy must be created in the surroundings - that is, at least −TΔS−TΔS energy must be lost as heat. This comes from the −ΔH−ΔH, and thus leaves −ΔH+TΔS−ΔH+TΔS to do useful work. No more work can be done than this. −ΔH+TΔS−ΔH+TΔS is −ΔG−ΔG, so −ΔG−ΔG is the maximum possible useful work. First of all, I wanted to know if this is correct, and if this is actually why −ΔG−ΔG is the maximum possible useful work.
Now if |ΔH||ΔH| is more than |TΔS||TΔS|, with both being negative, then one can think of the WmaxWmax or −ΔG−ΔG as a part of the |ΔH||ΔH|; since |ΔG|<|ΔH||ΔG|<|ΔH|. Some part of |ΔH||ΔH| goes as |TΔS||TΔS| to increase the surroundings' entropy, and the other part in doing useful work.
Hope it helps uuu
Similar questions