How do I solve for V in terms of p?

[psan6]

can somebody help me in solving this problem..thanks..

A gaseous substance whose properties are unknown, except as specified below, undergoes an internally reversible process during which

V = (-0.1p + 300)ft cube, when p is in psfa.

(a) for this process, find -/ V dp and /p dV, both in Btu, if the pressure changes from 1000psfa to 100psfa.

(/) means integral..thanks

 

[Jupiter6]

can somebody help me in solving this problem..thanks..

A gaseous substance whose properties are unknown, except as specified below, undergoes an internally reversible process during which

V = (-0.1p + 300)ft cube, when p is in psfa.

(a) for this process, find -/ V dp and /p dV, both in Btu, if the pressure changes from 1000psfa to 100psfa.

(/) means integral..thanks

This question doesn't make much sense but I suspect they're trying to impress upon you the importance of knowing what system your dealing with. Pdv is for closed systems. Vdp is for steady flow. You could try:
[tex]\int PdV = P_{avg}(V_2-V_1) [/tex]
and
[tex] \int -VdP = -V_{avg}(P_2-P_1)[/tex]
You can find V2 and V1 from your first equation.

 

[Mute]

What specifically is the problem? You don't know how to set up the problem or you don't know how to do the integrals? You have volume as a function of pressure - plug that expression into the integrals and integrate - of course, you'll have to solve the expression for pressure in terms of volume for one of the integrals. You're also given the initial and final pressure, so you can use that to find the intial and final volume.

For the integral with respect to pressure,

[tex]\int_{p_0}^{p_f}dp~V = \int_{p_0}^{p_f}dp~(-0.1p + 300)[/tex]

Can you integrate that?

For the other integral, express pressure as a function of volume, and perform the integral [itex]\int dV~p[/itex].