# Ideal gas formula not working?

ContrapuntoBrowniano
TL;DR Summary
Why does pressure and volume seem unnaffected for temperature changes in a heated cooking pot filled with water?
Hi! I wanted to do some basic calculations for temperature T on a water-filled pot. I noticed something strange on my calculations, and couldn’t figure out what was wrong...
So here it is:
The ideal gas formula:
k=PV
The actual formula Relates equally the product PV with the a constant associated with temperature. So, i never saw any flaws in this relationship for fluids...until today! You’ll see... If we vary the pressure or the volume leaving the other one constant, then the temperature will also vary accordingly, however, if we vary the temperature, the pressure and volume migth stay the same! This is the case of the heated water pot:
If water is an uncompressible fluid, the pressure changes appear insignificant to it, and since the only thing i’m doing is heating water, the volume stays the same...but “T” IS rising. Is there something I’m not considering?

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• ContrapuntoBrowniano
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This is the case of the heated water pot:
If water is an uncompressible fluid, the pressure changes appear insignificant to it, and since the only thing i’m doing is heating water, the volume stays the same...but “T” IS rising. Is there something I’m not considering?
Liquid water isn't a gas, so the gas laws do not apply to it.

• OmCheeto, russ_watters, robphy and 3 others
ContrapuntoBrowniano
Welcome!

The water molecules are accumulating energy that will end up in the liquid-steam phase change.

https://en.wikipedia.org/wiki/Enthalpy_of_vaporization

https://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html

If you freeze up the same water pot, you could measure an expansion or increasing volume.
Ok, but that does not account for this formula not working! Thermodynamics tells that
PV and nRT both have energy units (kg*m^2*s^-2 in SI) so: in this case n stays and so does R, but again, T changes?
Enthalpy does not come anywhere near this, and even if it does, the fluid is still changing i’ts temperature even when T<100C=212F
I’ve come up with some answers, but maybe they’re wrong:
-The formula only works for constant temperature.
-It only works for gases
-Infinitesimal variations in P or R result in considerable temperature changes for uncompressible fluids.

ContrapuntoBrowniano
Liquid water isn't a gas, so the gas laws do not apply to it.
That’s what i was thinking...however, this just moves the problem over to liquids.

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this just moves the problem over to liquids.
How? Liquids are not guesses.

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That’s what i was thinking...however, this just moves the problem over to liquids.
Wasn't the problem "the ideal gas law doesn't accurately describe the behavior of heating water" and the answer "liquid water isn't an ideal gas so the ideal gas law doesn't apply"? It seems to me the problem is solved.

So what problem are you referring to?

• vanhees71, OmCheeto, ContrapuntoBrowniano and 1 other person
Mentor
In general, it is important to know the limits of validity for any formula that you use. You cannot blindly plug in variables, you need to understand what those variables mean and when the formula applies to those variables. ##PV=nRT## applies only to ideal monoatomic gasses, and it is no longer valid when a material deviates significantly from the behavior of an ideal gas. That is the case with a liquid.

• vanhees71, jbriggs444, robphy and 3 others
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Ok, but that does not account for this formula not working! Thermodynamics tells that
PV and nRT both have energy units (kg*m^2*s^-2 in SI) so: in this case n stays and so does R, but again, T changes?
Enthalpy does not come anywhere near this, and even if it does, the fluid is still changing i’ts temperature even when T<100C=212F
I’ve come up with some answers, but maybe they’re wrong:
-The formula only works for constant temperature.
-It only works for gases
-Infinitesimal variations in P or R result in considerable temperature changes for uncompressible fluids.
https://en.wikipedia.org/wiki/Ideal_gas_law

- Isothermal thermodynamic processes are the only ones keeping temperature constant.

- Most real gases deviate from that ideal behavior.

- Variations of pressure do not result in considerable temperature changes for uncompressible fluids; take hydraulic fluid for example.

The nature of pressure in liquids (mainly due to surrounding pressure and gravity or internal molecular potential energy) is different than the internal pressure that naturally exists for liquids forced to remain within certain volume at certain temperature (or internal molecular kinetic energy).

• ContrapuntoBrowniano
ContrapuntoBrowniano
So what problem are you referring to?
How? Liquids are not *gasses
I’m referring to the problem of a Thermodynamic formula for the internal energy of liquids, in terms of heat capacity, amount and temperature. Maybe there is one? Something like an analogy for liquids of the Gas law.

• Lnewqban
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2022 Award
Summary: Why does pressure and volume seem unnaffected for temperature changes in a heated cooking pot filled with water?

Is there something I’m not considering?
What you are asking for is the equation of state. The ideal gas law is approximately valid for simple gases at moderate temperature and pressure. The surprise is not that is an inexact description for water but that it is as broadly applicable as it is. There are various approaches as indicated in the link.

• vanhees71 and ContrapuntoBrowniano
ContrapuntoBrowniano
What you are asking for is the equation of state. The ideal gas law is approximately valid for simple gases at moderate temperature and pressure. The surprise is not that is an inexact description for water but that it is as broadly applicable as it is. There are various approaches as indicated in the link.
Thanks! I’ll check it out. 😃🙌🏽

Gold Member
I’m referring to the problem of a Thermodynamic formula for the internal energy of liquids, in terms of heat capacity, amount and temperature. Maybe there is one? Something like an analogy for liquids of the Gas law.
A common approximation is to assume a constant volume process so dU=dQ. If evaporation becomes important, things get more complicated.

• Lnewqban
Gold Member
Summary: Why does pressure and volume seem unnaffected for temperature changes in a heated cooking pot filled with water?
Pressure rising along with temperature in a fixed volume is the basis of the pressure cooker. The higher pressure allows the temperature to get hotter than boiling water to cook faster.

https://scienceandsamosa.com/scienc...rks on,then pressure would naturally increase.

So, the Ideal Gas Law works fine you just have to put a lid on it!

• Dale and Lnewqban
Homework Helper
Gold Member
... Enthalpy does not come anywhere near this, and even if it does, the fluid is still changing i’ts temperature even when T<100C=212F
Copied from
https://en.wikipedia.org/wiki/Enthalpy

"The enthalpy H of a thermodynamic system is defined as the sum of its internal energy and the product of its pressure and volume:
H = U + pV
where U is the internal energy, p is pressure, and V is the volume of the system.

...
The U term is the energy of the system, and the pV term can be interpreted as the work that would be required to "make room" for the system if the pressure of the environment remained constant. When a system, for example, n moles of a gas of volume V at pressure p and temperature T, is created or brought to its present state from absolute zero, energy must be supplied equal to its internal energy U plus pV, where pV is the work done in pushing against the ambient (atmospheric) pressure."