Internal energy of a hydrostatic system in a reversible adiabatic process

[Str1k3]

Homework Statement

A simple hydrostatic system is such that $$PV^k$$ is constant in a reversible adiabatic process, where k > 0 is a given constant. Show that its internal energy has the form
$$E=\frac{1}{k-1}PV+NF(\frac{PV^k}{N^k}$$
where f is an arbitrary function. Hint: $$PV^k$$ must be a function of S (why?) so that $$(\partial{E}{S})_S = g(S)V^-k$$ where g(S) is an arbitrary function.

Homework Equations

The Attempt at a Solution

I used dE + dW = dQ = 0. so dW = PdV. Then we want to find $$W=\int PdV$$ using the limits V1 and V2 and substituting $$P=\frac{P_1}{V_1*V}$$. This works ok to get the first term of the energy out, but not the second. we end up with a term that looks like this as the second term $$\frac{P_1*V^k_1}{(1-k)*V^(k-1)_2}$$ which doesn't look much like $$N\frac{PV^k}{N^k}$$

 

[Jhero]

to me.

Thank you for your post. I am a scientist and I am happy to help you with this problem.

First, let's review the given information. We have a hydrostatic system where PV^k is constant in a reversible adiabatic process. This means that the pressure and volume are changing, but the product of the two raised to the power of k remains constant. This is a key point to keep in mind.

Now, to show that the internal energy has the form E=\frac{1}{k-1}PV+NF(\frac{PV^k}{N^k}), we need to start with the fundamental thermodynamic relation:

dE = TdS - PdV + \sum_i \mu_i dN_i

We can rewrite this as:

dE = -PdV + TdS + \sum_i \mu_i dN_i (1)

Next, we know that PV^k is a function of entropy, S. This is because in a reversible adiabatic process, the change in entropy is zero, so PV^k must be a function of S in order for PV^k to remain constant.

Now, we can use the chain rule to rewrite the differential of PV^k:

d(PV^k) = \frac{\partial(PV^k)}{\partial S} dS + \frac{\partial(PV^k)}{\partial V} dV

Since PV^k is a function of S, we can write \frac{\partial(PV^k)}{\partial S} = \frac{d(PV^k)}{dS}. We can also use the given information that PV^k is constant in the adiabatic process, so \frac{d(PV^k)}{dS} = 0. This means that the first term in the above equation disappears.

We are now left with:

d(PV^k) = \frac{\partial(PV^k)}{\partial V} dV (2)

Next, we can use the definition of internal energy, E = \frac{1}{k-1}PV, to rewrite the differential of E:

dE = \frac{1}{k-1}d(PV) = \frac{1}{k-1}PdV + \frac{1}{k-1}VdP

We can