Question

Write an equation for \[ d \rho \] in terms of \[ \rho \] and \[ dT \] if \[ \rho \] is a function of V and \[ dV = \beta V dT \]

Answer 1

For what it's worth, are all of these physical quantities of an ideal gas, or perhaps a solid?:
rho = density
T = temperature
V = volume

Answer 2

Yes that's the idea
Can you get \[ d \rho \] from \[d \rho/dt = d \rho/dV*dV/dt\]

Answer 3

but what is \[ d \rho/dV ?\]

Answer 4

oops I suppose those little t's above should be T's

Answer 5

By definition Mass = rho.V and as mass is constant, then
d(rho)/dV = d(mass/V)/dV = -mass/V^2

Answer 6

how would you get that in terms of \[ d \rho, \rho,\] and \[dT \]?

Answer 7

...so d(rho)/dV = -mass/V^2 = -mass/(mass/rho)^2 = -rho^2/mass

Now use the chain rule you have above and the fact that
dV/dT = beta.T

Answer 8

that would work great...but mass isn't specified in this problem.

Answer 9

no, but you can eliminate again with rho and V if you need to.

Answer 10

...I was sort of hoping by now you have seen that.

Answer 11

haha I just did as you were typing.

Answer 12

wait. so the units check out, but wouldn't that leave \[d \rho/dT = \beta T(- \rho^2/mass)\]

Answer 13

Right. Now if dV = beta.V dT, then you can solve for V in terms of T. Hence you can substitute into your equation above with V but then write it as V(T).

Then the only variables will be T and rho

Answer 14

you mean integrate?

Answer 15

i.e. dV/dT = beta.V ==> V = V_0.exp(beta.T)

Answer 16

yea. ya did. haha

Answer 17

Ok ... I'm off to bed. Good luck.

Answer 18

thank you sir