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Assume we have a balloon in a tank filled with a liquid , without any gravitational forces: We just know the Temperature, $V_0$ and $V_1$

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I want to calculate the pressure exerted from the liquid on the balloon. If it was just filled with gas I could easily calculate with $PV=nRT$, but this does not work for fluids.

Also by exclusion of gravity, the only source of pressure will be from the thermal energy ( fluid molecules moving and hitting on the balloon ) + internal cohesive energy.

UPDATE:

I think the only problem here is just finding the pressure exerted on the balloon by the liquid cohesive pressure ( in fact it is kind of equation of state problem ). I studied a lot since I asked this question. There are 2 pressures that I consider for using as P in this question:

1- Cohesive Energy density = 2.2973 GPa for water

2-Internal Cohesive Pressure= 168 MPa

My question is which one should be used here as $P$ ?

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    $\begingroup$ Can you clarify what you're asking? If you know $V_1$ the pressure in the ballon is just $P = nRT/V_1$, and assuming your system is in equilibrium the pressure in the liquid will be the same. The pressure in a liquid (or solid) is $P = K(V-V_0)/V_0$ where $V_0$ is the volume at zero pressure and $K$ is the bulk modulus. $\endgroup$
    – John Rennie
    Commented Mar 26, 2015 at 8:00
  • $\begingroup$ You might be interested in this post about the equation of state for liquids. $\endgroup$
    – Kyle Kanos
    Commented Mar 26, 2015 at 13:27
  • $\begingroup$ @JohnRennie - there will be a difference in the pressure in the balloon and the external liquid: this is due to the elastic forces in the balloon. $\endgroup$
    – Floris
    Commented Mar 31, 2015 at 22:28
  • $\begingroup$ Two questions: (1) Do you know n (number of air molecules)? (2) Is the balloon made of a stretchy elastic material like rubber? (If so, the rubber provides almost all the pressure and the water outside can be ignored!) $\endgroup$
    – Steve Byrnes
    Commented Apr 6, 2015 at 19:17
  • $\begingroup$ @SteveB no but we know the volume of the balloon. just ignore the balloon material. Actually it is just an air cavity made inside the water with no boundary material. $\endgroup$
    – Aug
    Commented Apr 6, 2015 at 19:21

2 Answers 2

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Considering your comment "Actually it is just an air cavity made inside the water with no boundary material":

$P_{in} = P_{out} + 2\gamma /R$

where $R$ is the radius of the air cavity and $\gamma$ is the surface tension of the air-water interface, which is 0.072 N/m at 298K.

See Internation Tables of the Surface Tension of Water for values at other temperatures.

To calculate $P_{out}$ as a function of T and V, you need an equation of state for liquid water.

See A new analytic equation of state for liquid water J. Chem. Phys., Vol. 110, pages 484-496.

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You say you know the temperature and the volume of the water. Assuming that you also know the mass of the water, you know both temperature and specific volume (the inverse of the the density). You just need a thermodynamic table for the properties of liquid water like this one from NIST to find the pressure. If you actually don't have enough information to find the mass of the water, then you don't have enough information to find the pressure.

Compressed Liquid tables are normally indexed by $P$ then by $T$- i.e., there are sub-tables for different $P$ values and each sub-table is indexed by $T$. The first italicized row in the linked table tells you that if you had $T=0$ $^\circ$C then $\rho=999.8$ kg/m3 would indicate $P = 0.01$ MPa (from the 2nd column, which is in the 0.01 MPa sub-table) or $0.02$ MPa (from the seventh column).

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