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please help with partial differentiation

Hi, not sure if this is in the right forum because these are first year degree qus, but if anyone can help would really appreciate it:

1) for (xdy-ydx)(x^2+y^2)
firstly had to determine that this was exact, if in the form Pdx+Qdy then
(dP/dy)=(dQ/dx)=(y^2-x^2)/((x^2+y^2)^2)

i)how do i find a function f such that df=Pdx+Qdy?
ii)what is the general solution to the equation Pdx+Qdy=0 im guessing this will be f=constant but just dont know how to get f.

2)if enthalpy of a gas is H=U+pV where U satisfies dU=TdS-pdV then what is the relationship between the variables H,S and p? how do i show that (dV/dS)with p constant is equal to (dT/dP) with s constant?

By regarding U as a function of p and V and considering 2 expressions for d2U/dpdV show that:

(dS/dV)(dT/dp)-(dS/dp)(dT/dV)=1
p v v p (these letters are constants for each differential)

thanks in advance for any suggestions x

Reply 1

RichE

15

Jitterbug

Hi, not sure if this is in the right forum because these are first year degree qus, but if anyone can help would really appreciate it:

1) for (xdy-ydx)/(x^2+y^2)
firstly had to determine that this was exact, if in the form Pdx+Qdy then
(dP/dy)=(dQ/dx)=(y^2-x^2)/((x^2+y^2)^2)

i)how do i find a function f such that df=Pdx+Qdy?
ii)what is the general solution to the equation Pdx+Qdy=0 im guessing this will be f=constant but just dont know how to get f.

You're right in two, the general soln will be f(x,y)=c where f is such that

df/dx = P and df/dy = Q.

Now we know that

df/dx = -y/(x^2+y^2).

If we integrate wrt x then we get

f(x,y) = -arctan(x/y) + g(y).

Note that we have a function g of y, instead of the usual constant, because this would disappear when we partially differentiated wrt x.

If we partial-differentiate this equation wrt y we get

df/dy = -x/(x^2+y^2) + g'(y)

which is what we want if g'(y)=0. i.e. if g is a constant.

So we can take f(x,y) = arctan(x/y) and the general solution is

arctan(x/y) = c.

Reply 2

RichE

15

Can you give any more information on the second one?

You have that H is a function of U,p,V and the second equation gives U in terms of T,S,p,V so I can't see that you can get H in terms of just S and p. :confused:

Or are you looking for a differential equation relating H,S,p?

Reply 3

Jitterbug

OP

RichE

Can you give any more information on the second one?

You have that H is a function of U,p,V and the second equation gives U in terms of T,S,p,V so I can't see that you can get H in terms of just S and p. :confused:

Or are you looking for a differential equation relating H,S,p?

i think its just a differential equation relating H S and p. its something to do with thermodynamics and maxwell identities but not sure how to go about it.

Reply 4

RichE

15

But does it just involve H, s and p? How do you eliminate the other variables? Are there other conservation relating the other variables that I need to know about?

Reply 5

Fermat

8

RichE

But does it just involve H, s and p? How do you eliminate the other variables? Are there other conservation relating the other variables that I need to know about?

The gas eqn is pV = mRT (mR = const)
can you use that to eliminate T?

Reply 6

nas7232

19

pv = nRT

Therefore, pv = (m/M) RT

You cant eliminate T like that i dont think.

BTW i didnt read the thread i just saw the equation, this idea gas equation is in my physics syllabus.

Reply 7

evariste_3086

1

Jitterbug

.
2)if enthalpy of a gas is H=U+pV where U satisfies dU=TdS-pdV then what is the relationship between the variables H,S and p? how do i show that (dV/dS)with p constant is equal to (dT/dP) with s constant?

H=U+pV
dH=dU+d(pV)
=TdS-pdV+vdp+pdv
=TdS+vdP.......(1)
so H=H(S,P)
let D denote partial diff
from (1)
DH/DS=T DH/Dp=v
now D^2H/DSDP=Dv/DS=D^2H/DpDS=DT/DP
which gives Dv/DS=DT/Dp.
probably same idea for second part. ill look later if i get time, got some lesson plans to write up now.

Reply 8

RichE

15

evariste

H=U+pV
dH=dU+d(pV)
=TdS-pdV+vdp+pdv
=TdS+vdP.......(1)
so H=H(S,P) (*)
let D denote partial diff
from (1)
DH/DS=T DH/Dp=v
now D^2H/DSDP=Dv/DS=D^2H/DpDS=DT/DP
which gives Dv/DS=DT/Dp.
probably same idea for second part. ill look later if i get time, got some lesson plans to write up now.

How is (*) a relationship between H,S,when T and v are also involved?

Reply 9

evariste_3086

1

RichE

How is (*) a relationship between H,S,when T and v are also involved?

Because from (1) small changes in S and p produce small change in H. ie H varies as S and p do. it is (probably?) possible to define H in terms of other variables too.

please help with partial differentiation

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