<p>Now, we all know that during phase change, when heat is supplied, only one of kinetic energy or potential energy change. My question is, in the graph above, where is change in kinetic/potential energy indicated?</p>
<p>When the slope is equal to 0, is there change in kinetic energy? Or Potential Energy?</p>
<p>Obviously, when the slope is not equal to zero, change in the other type of energy will be present. Thanks for answering!</p>
<p>From what I remember of chem the y axis is actually the temperature of the substance and you can look at the x axis as the energy being put into the system. The reason for the slopes of zero is because, although energy is being inputted, the temperature of the substance is not increasing. </p>
<p>You can consider the y axis (temperature) to be kinetic energy. The x-axis is more like total energy added. If the two axis have the same units (I dont remember if they do) then you can subtract the x axis from the y to see the potential energy.</p>
<p>I’m just saying what makes sense to me. I could be wrong on the specifics.</p>
<p>Robotnik, you are correct in saying that I can take the y axis to be kinetic energy (because the higher the temperature of a substance, the higher will be the average kinetic energy of the molecules). And you are also correct in taking the x axis as total energy required (because an increase in temperature is only due to supply of energy, and so according to the situation you are suggesting, the x axis can be used to find out the total energy required to reach a particular kinetic energy of molecules).</p>
<p>Though I am not too sure about the last part you are suggesting “then you can subtract the x coordinate from the y coordinate to get the potential energy”. I see three things wrong with this:</p>
<ol>
<li>If this statement is true, then it should be “subtract the y coordinate from the x coordinate to get the potential energy” and not the other way round.</li>
<li>I doubt that this statement is based on an actual concept.</li>
<li>Whenever heat is supplied to a substance, either only kinetic energy changes or only potential energy changes. What you are suggesting implies that at a point where there is change in kinetic energy (on the slope), there is a change in potential energy as well. Which of course is contradicting the original concept.</li>
</ol>
<p>But, my question is still not answered. Let me try to put it in mathematical terms:</p>
<p>Q. Does the part of the graph where m=0 depict change in kinetic energy or change in potential energy of the substance?</p>
<p>PS. Now that I come to think of it… for the 3rd point I wrote, I think the parts of the graph where m=0 depicts change in potential energy. And the slanting parts of the graph depict change in kinetic energy. Could someone please concur? Thanks!</p>
<p>Your first point is correct. My mistake. I am replying on my iPhone so looking back at previous responses is difficult and things get mixed up. </p>
<p>As to your second point: yes, I am not basing my response on anything official. I am simply saying what makes sense to me. </p>
<p>As to your final statement about when the graph platos it is because it is a change in PE rather than KE: that’s is exactly what I was trying to convey. My apologies if that was unclear.</p>
<p>Yes! I think I’ve got it, but I am still not sure. Here’s my explanation:</p>
<ol>
<li>At m=0, the change in temperature is 0.</li>
<li>Change in temperature changes the average kinetic energy of molecules.</li>
<li>Therefore, there is no change in the average kinetic energy of the molecules.</li>
<li>At any point in the graph, there can only be change in kinetic energy or change in potential energy.</li>
<li>At no point, is there no change in the energy.</li>
<li>Therefore, at m=0, there is change in potential energy.</li>
</ol>
<p>Conclusion: At the horizontal portions of the graph, there is change in potential energy. At the slanting parts, there is change in kinetic energy.</p>
<p>Also, if anyone else has any input, could they also tell me if my line of reasoning is correct?</p>
<p>Temperature of an ideal gas is essentially kinetic energy in different units. The x axis is total energy inputted (PE+KE). If temp. is converted to KE, couldn’t you figure out the exact value of PE by subtracting KE from total energy?</p>
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