Fluid Mechanics - Water, Ridiculous Answer

[eurekameh]

Water at standard atmospheric pressure and temperature fills a sealed, rigid container. If the temperature of the water is increased by 50°C, what is the pressure?

V = volume, P = pressure, T = temperature

Coefficient of compressibility, kappa = (-1/V)(dV/dP)
Coefficient of thermal expansion, beta = (1/V)(dV/dT)

I solved for dV/V from the coefficient of thermal expansion equation:
dV/V = beta*dT.

Plugging into coefficient of compressibility equation:
kappa = -(beta*dT)/dP.

Solved for dP = -(beta*dT)/kappa, changed the d's into deltas.

Plugging kappa = 5.18 * 10^-10 and beta = 1.48 * 10^-4 into delta P equation gave me delta P = -92.3 MPa. Is this reasonable at all? I mean, the pressure difference seems a bit large.

 

[eurekameh]

Anyone?

 

[TimSmith]

Hi eurekameh,

Now without even looking through your equations I can tell you that they are wrong, or you have made a mistake somewhere.

An increase in temperature will cause an increase in pressure, vice versa... Look up how refrigerators and heat pumps work, also think of your deodorant can - it loses pressure and gets colder.

I'm afraid I can't help you much more as my brain is currently fried from work.

 

[Chestermiller]

Water at standard atmospheric pressure and temperature fills a sealed, rigid container. If the temperature of the water is increased by 50°C, what is the pressure?

V = volume, P = pressure, T = temperature

Coefficient of compressibility, kappa = (-1/V)(dV/dP)
Coefficient of thermal expansion, beta = (1/V)(dV/dT)

I solved for dV/V from the coefficient of thermal expansion equation:
dV/V = beta*dT.

Plugging into coefficient of compressibility equation:
kappa = -(beta*dT)/dP.

Solved for dP = -(beta*dT)/kappa, changed the d's into deltas.

Plugging kappa = 5.18 * 10^-10 and beta = 1.48 * 10^-4 into delta P equation gave me delta P = -92.3 MPa. Is this reasonable at all? I mean, the pressure difference seems a bit large.

The magnitude of your answer is correct, but the sign is wrong. Write the equation of the change in V in terms of the changes in P and T. Then set the change in V equal to zero in the equation.

 

[rezeew]

I would say that this answer is not accurate or reasonable. The pressure difference of -92.3 MPa is unrealistic and would not occur in a sealed, rigid container with water at standard atmospheric pressure and temperature. The equations used in this answer are valid for ideal conditions, but in reality, there are many other factors that would affect the pressure, such as the rigidity of the container and the compressibility of water itself. A more realistic approach would be to use the ideal gas law or other relevant equations that take into account these factors. It is important to critically evaluate and question any results or answers that seem unreasonable or unrealistic in the field of science.