what is ratio of pressure to water temperature in closed container?

Question

I am sure to a lot of people this is very simple question but to me I do not know the ratio,Thank you in for any info on this.

Answer 1

I know that, for a gas:

PV = NkT

where P is the pressure, V is the volume, N is the number of moles, k is plancks constant and T is the temperature, so the ratio of pressure to temperature for a gas of constant volume is:

P/T = Nk/V

I do not, however, know whether this holds true for a liquid.

Answer 2

It depends. If you have a pocket of gas then as you heat the water and the container and the gas, the container will probably expand a little ( partly due to thermal expansion partly due to the increase in pressure elastically stretching the container sides ) the gas will tend to compress as the water will most likely expand more than the container. The increase in pressure in the gas will depend on the increase in temperature in the gas, the reduction in volume in the gas and on whether the gas is absorbed more in the warm water, also on the increased water vapour pressure in the pocket.

If you have no gas and the container material's coefficient of thermal expansion is much less than water then the increase in pressure will depend on elastic strain in the container.

Answer 3

Since water is incompressible, I don't think there is a significant ratio. I googled for this and got an answer that seems reasonable:
"Can anyone explain in plain English what the pressure rise of water will be in a closed vessel with no entrained air per degree Kelvin rise in temperature? I'm not concerned about nonlinearities because the temperature range is 290oK to 305oK (17oC to 32oC or 60oF to 90oF). "

Thermodynamically, for liquid water there will be no pressure change related to the change in temperature for that range. a good model for this reasoning is dP/dT|v = αp/kT. for water: αp = 207K-1 and kT = 49000000bar-1

solving for dP gives dP =(αp/kT)*dT. since αp/kT is such a small term (4.22*10-5) you would need a very large dT to see a change in dP.

so that rules out the temperature dependency for pressure based on having only liquid water in your system. Im still thinking about the other scenerio with entrainment.