Adiabatic expansion on $(P,T)$ diagram

Question

Consider a certain substance which is described by pressure $P$, volume $V$ and temperature $T$. In one thermodynamics exercise I was asked to draw one adiabatic expansion from $V_1$ to $V_2$ in a $(P,T)$ diagram.

Now, I'm wondering, with just this little information, can someone do it? I mean, adiabatic only means that heat is not exchanged in the process. Also, since no one said which substance that is, we don't have any relation between $P,V,T$ at first. In that case we don't even know how the states $(P_1,T_1)$ and $(P_2,T_2)$ are positioned, let alone know the form of one adiabatic curve for this system.

In that case, is there any way to draw one adiabatic expansion with so little knowledge of the system?

I mean if it were an ideal gas we know that adibatic processes obey $PV^\gamma = k$ where $k$ is a constant. In that case, since

$$PV = NRT,$$

we know that $V = NRT/P$, so that

$$P(NRT/P)^\gamma = k \Longrightarrow P^{1-\gamma}T^\gamma=k'$$

where $k'$ is another constant. The whole point is: we can know the form of adibatic processess in the $(P,T)$ plane because we have information about the gas.

But in this case, we have nothing. It is just said it is a substance, but it could be anything.

In that case, how could we draw a diagram for an adiabatic process?

Answer 1

For a general single-phase constant-composition substance, dU is expressible as:

$$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV$$ So, for an adiabatic reversible expansion, $$dU=C_vdT+\left[T\left(\frac{\partial P}{\partial T}\right)_V-P\right]dV=-PdV$$ Therefore, we have $$C_vdT=-T\left(\frac{\partial P}{\partial T}\right)_VdV\tag{1}$$ Also, from the equation of state: $$dV=\left(\frac{\partial V}{\partial T}\right)_pdT+\left(\frac{\partial V}{\partial P}\right)_TdP\tag{2}$$ Combining Eqns. 1 and 2, we obtain: $$C_vdT=-T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_pdT-T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial P}\right)_TdP\tag{3}$$or $$\frac{dT}{dP}=-\frac{T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial P}\right)_T}{C_v+T\left(\frac{\partial P}{\partial T}\right)_V\left(\frac{\partial V}{\partial T}\right)_p}\tag{4}$$ The term in the numerator can be simplified further such that $$\frac{dT}{dP}=\frac{T\left(\frac{\partial V}{\partial T}\right)_P}{C_v+T\left(\frac{\partial V}{\partial T}\right)_p\left(\frac{\partial P}{\partial T}\right)_V}\tag{5}$$ Eqn. 5 reduces to the appropriate relationship for an ideal gas. All the terms in this equation are positive.

Answer 2

From the ideal gas law you know that $ P = \frac{T}{V} n R $ and so the adiabatic process is solved for temperature $$T = \frac{V^{1-\gamma}}{\frac{n R}{k}}$$

and also the pressure $$ P = k V^{-\gamma}$$

$$(T,P) = ( \frac{V^{1-\gamma}}{\frac{n R}{k}}, k V^{-\gamma} )$$

So the only information we need is the initial conditions (the give us $k$) and the ratio $\gamma = c_p/c_v$. The reason the chemical composition or material properties do not play a role is that in a (ideal) gas the molecules are far apart not to interact with each other. Are you familiar with the kinetic theory of gasses? Essentially they are like little spheres bouncing off the boundary (walls) causing pressure from the exchange in momentum.