n₁ moles of a monatomic gas and n₂ moles of a diatomic gas are mixed together in a container. a. Derive an expression for the molar specific heat at constant volume of the mixture.

Video transcript

Hi, everyone. In this practice problem, we're being asked to find an expression that can be used to calculate specific heat at constant volume. We'll have two poly atomic gasses that are mixed together. And the composition of the gasses is as follows. Yes, one will have N one moles and degrees of freedom F equals to seven and yes, two will have N two moles and degrees of freedom F equals to nine. We're being asked to find an expression that can be used to calculate the specific heat at constant volume for one mole of guess the options given listing the different expressions from A to D that can be used to calculate the specific heat at constant volume or CV. So we want to recall that four poly atomic gasses CP can be calculated by CP equals to half multiplied by F multiplied by R. So in this case, for guess one CV, one, the F is going to be seven. So we have CV, one then be equal to seven divided by two multiplied by R. Similarly CV two will have an, an F of nine. So the CV two will then becomes nine divided by two multiplied by R. We can then look into the thermal energy for the poly atomic gasses. The first one is going to be E one or in this case, the F is going to be equal to seven. So E one will then be equal to half multiplied by seven multiplied by R multiplied by N one delta T simplifying this E one will then come out to B seven divided by two multiplied by R and one delta T. For the second gas or E two, we will follow the same formula where E two equals to half multiplied by nine, multiplied by R multiplied by N two multiplied by delta T which will then be simplified into nine divided by two multiplied by RN two delta T Awesome. We now get the total energy of the mixture, which can be calculated using this E DH formula which will be equal to E one plus E two, which in this case, then we can calculate this to be equal to half multiplied by, in parentheses, seven multiplied by N one plus nine multiplied by N two multiplied by R multiplied by delta T. So there our um eth, since we are looking at the delta D will come out to be delta EPH and simplifying our formula. Delta eth will then come out to be half multiplied by, in parentheses, seven N one plus nine N two multiplied by R multiplied by delta T. We can then compare the equation with delta eth or the equation for delta EPH, that we usually know which is N total multiplied by CP multiplied by delta T. Or in this case, that will be equal to N one plus N two multiplied by CP multiplied by delta T. So N one plus N two multiplied by CV, multiplied by delta T will actually be equal to the delta eth expression that we have obtained previously by summit uh summit, taking the summation of E one and E two. So in this case, I want to equate the two. So N one plus N two multiplied by CV, multiplied by delta T will actually be equal to half multiplied by seven N one plus nine N two multiplied by R multiplied by delta T. So what we wanna do next is to just uh cross out the delta T and then from there, we want to rearrange our equation so that we will get an equation for CV, which will then come out to B in parentheses seven N one plus nine N two multiplied by R define it by two, multiplied by N one plus N two. And this will be the final expression for CV, which will be the final answer to this particular practice problem, which will actually correspond to option A in our answer choices with CV equals two in parentheses seven and one plus nine and two close parentheses multiplied by R divided by two in parentheses. And one plus N two. So option A will be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website or other than that, that'll be it for this video and thank you so much for watching. Bye bye.