Question

calculate the work done during the isothermal reversible expansion of the van der waals gas.account physically for the way in which the coefficient a and b appear in the final expression. plot the same graph the indicator diagrams for the isothermal reversible expansion of
a)a perfect gas
b)a van der waals gas in which a=0,b=5.11*10^-2 dm^3/mol
c)a van der waals gas in which a=4.2 dm^6 atm/mol,b=0.
the values se;ected exaggerate the imperfections but give rise to significant effects on the indicator diagrams. V=1 dm3,n=1 mol,T=298K.

Answer 1

I think you can post this in Chemistry section too.

Answer 2

Let's remember that \[W=\int\limits_{}^{}Fdx\] So this tells us work is the integral of force times the distance.

But for a gas we need to think about pressures, so we take our definition of pressure and rearrange it.\[p=\frac{ F }{ A }\] then substitute it in... \[W=\int\limits_{}^{}pAdx\]Then we can see we have A and dx multiplied together. A is a meter*meter, or a squared meter, and dx is an infinitesimally small distance in units of meters, so we can say that

Adx=dV

Which is just showing that dV (change in volume) is proportional to a change in area times another length. This might sound confusing, and if it is I'll explain it better, but it might just make sense.

So we substitute that in and get this:\[W=-\int\limits_{}^{}pdV\] we're almost done, but we need to remember that this is from the perspective of the environment. That means when the volume decreases, the environment is doing work on the system and internal energy of the system increases. Conversely, if the system expands, it's pushing at the environment and in doing so loses internal energy. So in order to account for this, we need to put a negative sign in front of the integral.

This is not something you can understand in a few seconds, you will need to think about it for a while and develop a model in your mind to understand why and read more. I can help you only so much, but I am happy to help explain things as best I can to give you the ability to think with these fundamental thermodynamic concepts that all good physical chemists think with when designing experiments and solving problems.