Finding coefficient of thermal expansion and isothermal compressibility

[osker246]

Homework Statement

To a very good approximation, ammonia obeys the Bertholet equation of state,
which readsPV=nRT+\frac{9}{128}(\frac{nRTc}{Pc})(1-6\frac{Tc^2}{T^2})Pa)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.

Homework Equations

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 I am 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.

 

[SteamKing]

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?

 

[osker246]

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