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Mathematics 15 Online
OpenStudy (anonymous):

DIFFERENTIAL EQUATION: The ideal gas law relating pressure ,temperature and volume is given by: P =(CT)/V where c is a constant. Find TdP/dT * dV/dT

OpenStudy (anonymous):

\[T \frac{ \delta P }{ \delta T } *\frac{ \delta V }{ \delta T }\]

OpenStudy (anonymous):

A hint: once write P in terms of T and then differentiate, another time write V in term of T and then differentiate.

OpenStudy (anonymous):

P in terms of T:\[P= \frac{ CT }{ V}\] and \[\frac{ \delta P }{ \delta T } = C/V ?\]

OpenStudy (anonymous):

first differentiate P w.r.t T assuming V constant after that differentiate V w.r.t. T as P constant. then put in the equation.

OpenStudy (theeric):

@MERTICH , I'm not sure, but you might be interested in using the \(\LaTeX\) symbol \(\partial\) (\partial). That's what I use, but maybe you meant to use \(\delta\)! I just know that \(\LaTeX\) isn't common knowledge.

OpenStudy (theeric):

@MERTICH Looks good so far!

OpenStudy (theeric):

Take care, all.

OpenStudy (anonymous):

thanks, let me take care of the other one

OpenStudy (anonymous):

Yes, you did OK. try to do the second one.

OpenStudy (anonymous):

\[V = \frac{ CT }{ P }, \frac{ \delta V }{ \delta T } = \frac{ C }{ P }\]

OpenStudy (anonymous):

Yep, you have done all OK.

OpenStudy (anonymous):

thank you @Saeeddiscover , now, do i just need to multiply them?

OpenStudy (anonymous):

there is a dot . between them

OpenStudy (anonymous):

Yes, it is what you need to do.

OpenStudy (anonymous):

thank you!

OpenStudy (anonymous):

No problem!

OpenStudy (anonymous):

so what happpens with the T ,\[T \frac{ \delta P }{ \delta T} *\frac{ \delta V }{ \delta T}\], WHEN I MULTIPLY, is that T at first included?

OpenStudy (theeric):

Hi! I notice that you haven't quite finished! You do multiply. If you substitute in the partial derivatives that you found, you're left with \(T\left(\dfrac C V\right)\dot\ \left(\dfrac C P \right)\) This can be simplified, so you should. \(\dfrac {T\ C} V\dot\ \dfrac C P=\dfrac {C\ T} V\dot\ \dfrac C P\) That was a very slight algebraic move there.. I know you're not on but would like to think about this if you were here. Does that \(\dfrac{CT} V\) look familiar? \(\\\downarrow\\\downarrow\\\downarrow\) It is \(P\)! Look at the pressure equation from the beginning. So we have \(P\ \dot\ \dfrac C P\) That is, \(T \dfrac{ \delta P }{ \delta T} *\dfrac{ \delta V }{ \delta T}=P\ \dot\ \dfrac C P=C\) And then \(that\) would be your simplified solution.

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