# Finding coefficient of thermal expansion and isothermal compressibility

1. Jul 16, 2011

### osker246

1. The problem statement, all variables and given/known data

To a very good approximation, ammonia obeys the Bertholet equation of state,

PV=nRT+$\frac{9}{128}$($\frac{nRTc}{Pc}$)(1-6$\frac{Tc^2}{T^2}$)P

a)Suppose we have 500 grams of ammonia under a pressure of P=3.04 atm
and at T=323K. Calculate the volume of ammonia according to the
Bertholet equation of state and compare to the result predicted by the ideal
gas law.

b)Assuming ammonia obeys the Bertholet equation of state obtain
expressions for the coefficient of thermal expansion$\beta$=$\frac{1}{V}$($\frac{dV}{dT}$)p and the isothermal compressibility $\kappa$=$\frac{-1}{V}$($\frac{dV}{dP}$)T (note: these are partial derivatives at constant P and T). Evaluate β and κ for 500 grams of ammonia at P=3.04 atm and at T=323K.

c)Using your results from part b, calculate ($\frac{dU}{dV}$)T and ($\frac{dH}{dP}$)T for 500 grams of ammonia at P-3.04 atm and T=323K.

2. Relevant equations

3. The attempt at a solution

Ok, so I found the answer to part A which was 0.251 m^3 using Bertholet eqn. of state and 0.256 m^3 using ideal gas law.

Now im not sure about part B. I have a feeling I can accomplish this buy simply solving for volume with Bertholet eqn. of state (or ideal gas law) and simply evaluating the derivative at that point; with T being my variable for beta and P being the variable for kappa. Is that the proper way to evaluate beta and kappa in this situation? Thanks for the help.

Last edited: Jul 16, 2011
2. Jul 16, 2011

### SteamKing

Staff Emeritus
There is something wrong with your units.

Using PV = nRT, I calculate V = 0.256 cubic meters for 500 grams of NH3 at 3.04 atm and 323 K. Have you used the correct R value?

3. Jul 16, 2011

### osker246

Ahh yes I forgot to convert pressure into Pascals. Thanks for the heads up.