[vivekrai]

At 298K, the thermal expansion coefficient and the Isothermal compressibility of liquid water are and respectively. Then, for water at 300 K and P = 1.00 bar is :

*a) 1.5 * 10^4 bar
b) 3 * 10^-4 bar
c) 6 * 10^-4 bar
d) 9 * 10^-4 bar

I don't know how to proceed after this $$\partial \text{U}=\text{T}\partial S-\text{P}\partial V$$

[juanrga]

At 298K, the thermal expansion coefficient and the Isothermal compressibility of liquid water are and respectively. Then, for water at 300 K and P = 1.00 bar is :

*a) 1.5 * 10^4 bar
b) 3 * 10^-4 bar
c) 6 * 10^-4 bar
d) 9 * 10^-4 bar

I don't know how to proceed after this $$\partial \text{U}=\text{T}\partial S-\text{P}\partial V$$

In ## dU = T dS - P dV ##  you have a function ## U = U(S,V) ##, but in your above partial derivative

$$ \left( \frac{\partial U}{\partial V} \right)_T$$

you need a function ## U = U(T,V) ##.

You have to use the correct function if you want compute the partial derivative.

[Enthalpy]

4*10^-6 bar can't be correct in your data!

Bulk modulus (adiabatic) for liquids and polymers is between 0.7GPa and 3.5GPa around 1atm (some liquid metals exceed that), with water around 2.5GPa if memory serves. That's 25,000 bar or 40*10^-6 bar^-1 adiabatically.

That is, you need many bar to change significantly the volume.

[vivekrai]

Will this : $$ \left(\frac{\partial U}{\partial V}\right)_V =T\left(\frac{\partial p}{\partial V}\right)_T - P$$ ?

[juanrga]

4*10^-6 bar can't be correct in your data!

Bulk modulus (adiabatic) for liquids and polymers is between 0.7GPa and 3.5GPa around 1atm (some liquid metals exceed that), with water around 2.5GPa if memory serves. That's 25,000 bar or 40*10^-6 bar^-1 adiabatically.

That is, you need many bar to change significantly the volume.

You are right. The isothermal compressibility coefficient (κ) of water at 50 ºC and 1 bar is about 44 x 10^-6 bar^-1.

[juanrga]

Will this : $$ \left(\frac{\partial U}{\partial V}\right)_V =T\left(\frac{\partial p}{\partial V}\right)_T - P$$ ?

How you take the partial derivative against volume maintaining volume constant? Why two different Ps?...

[vivekrai]

Will this : $$ \left(\frac{\partial U}{\partial V}\right)_V =T\left(\frac{\partial p}{\partial V}\right)_T - P$$ ?

How you take the partial derivative against volume maintaining volume constant? Why two different Ps?...

That's just Latex gone wrong. Also may be because I haven't been into thermodynamics so much. Should I leave this because I have not read this? Or I should get some information on it(How?)

[juanrga]

That's just Latex gone wrong. Also may be because I haven't been into thermodynamics so much. Should I leave this because I have not read this? Or I should get some information on it(How?)

Ok I assume that you have the correct Helmholtz equation
$$\left( \frac{\partial U}{\partial V} \right)_T = \left[ T \left( \frac{\partial p}{\partial T} \right)_V - p \right]$$
The right hand side is a function of pressure, temperature, and isocoric variation of pressure with temperature. Pressure and temperature are given and you would be able to compute the isocoric variation of pressure with temperature. Do you know the definitions for ##\beta## and ##\kappa## and how to do this?