lecturenotes_nov182010

lecturenotes_nov182010 - Lecture notes for Lecture #8,...

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Lecture notes for Lecture #8, November 18, 2010 Molar Heat Capacities From equipartition we should now be very familiar with the following equation: ± ²³´µ¶·¸± ¹ º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ Æ Ç È É Now for a given number of active modes, everything on the right side of the equals sign is a constant with the exception of the temperature, T, so we can safely assume that any change in thermal energy is going to be related to a change in temperature. This allows us to rewrite the equation above defined over an interval of changing temperature: Ê ± ²³´µ¶·¸± ¹ º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ Æ Ç È ÊÉ And we know from the very first week of the quarter that thermal energy changes are directly related to changes in temperature where the heat capacity is the constant of proportionality. This allows us to write Ê ± ²³´µ¶·¸± ¹ ËÊÉ If we equate these two expressions for Ê ± ²³´µ¶·¸± ±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±±º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ Ì Í Î ÊÉ ¹ ËÊÉ º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ Ì Í Î ¹ Ë It is important at this point to breakdown the item in parenthesis, º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ , into something useable, perhaps number of modes/atom as a start. º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ = º¿Ïб¶ÏÑ´ÒÅ ·²Ï¶ Ó ºÔÕÁÖÃ×±¼À±½»¼ÁÄÅ So let us say that I gave you 50 atoms of a monotomic gas and I asked how many modes were available, here is how we would find the answer: For a monotomic gas, the º¿Ïб¶ÏÑ´ÒÅ ·²Ï¶ is 3, and we have a total of 50 atoms. Therefore: º»¼»½¾±¿±¼À±Á¼ÂÃÄÅ = 3 * 50 = 150
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Now suppose instead of 50 atoms, I gave you 4 dozen atoms. Here we will employ some minor unit analysis that there are 12 atoms in a dozen. Again for a monotomic gas, the ±²³´µ²¶·¸¹ º»²µ is 3, and we have a total of 4 dozen atoms. Therefore: ¼½¼¾¿´±´½À´Á½ÂÃĹ = ±²³´µ²¶·¸¹ º»²µ Å ÆÇ´´º»²µ¸¹ ¶²È·É Å ÊËÁÌÃÍ´½À´Â½ÎÃÊĹ ¼½¼¾¿´±´½À´Á½ÂÃĹ Ï (3) (12) (4) = 144 Now suppose instead of 4 dozen atoms, I gave you 4 moles of atoms. Here we will employ some additional unit analysis there are 6.02 X 10 23 atoms in a mole. Once more, for a monotomic gas, the ±²³´µ²¶·¸¹ º»²µ is 3, and we have a total of 4 moles of atoms. Therefore: ¼½¼¾¿´±´½À´Á½ÂÃĹ = ±²³´µ²¶·¸¹ º»²µ Å ´ÐÑÒÇ´Ó´ÆÒ ÔÕ º»²µ¸¹¹ µ²Ö· Å ×Á½¿ÃĹ ¼½¼¾¿´±´½À´Á½ÂÃĹ Ï (3) ( 6.02 X 10 23 ) (4) Or more generally ¼½¼¾¿´±´½À´Á½ÂÃĹ Ï ±²³´µ²¶·¸¹ º»²µ ( Ø º ) (# of moles) Now let us remember that ¼½¼¾¿´±´½À´Á½ÂÃĹ Ù Ú Ç Ï Û , so if we take the expression above that we found for ¼½¼¾¿´±´½À´Á½ÂÃÄ¹Ü and replace it in the expression for heat capacity C, we arrive at: Û Ï ¼½¼¾¿´±´½À´Á½ÂÃĹ Ù Ú Ç Ï
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lecturenotes_nov182010 - Lecture notes for Lecture #8,...

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