Thermodynamics of capacitor filled with a linear ideal gas dielectric

[frmn]

Homework Statement

Consider a long cylindrical vessel of base area A_0. The ends of the vessel are fitted with two massless metallic plates also of area A_0. The bottom plate is fixed while the top plate is free to move. The initial length of the cylindrical column between the plates is L_0, in which n moles of an ideal gas (f=4) are filled, initially at a temperature T_0. It has a hypothetical permittivity ε(T)=k/T, and acts as an isotropic dielectric. The walls of the vessel are electrically insulating but allow heat exchange. The two metallic ends are connected to the terminals of an ideal battery of constant emf V_0. Initially, the upper plate is in equilibrium. The gas is then subjected to a polytropic process such that the process is quasi static, so as to expand until the length of the column is 2L_0. Thus, find the polytropic index a of the process, and also the heat exchange (if any).

Relevant Equations

PV=nRT
dU=TdS-dW
P_elec=σ²/2ε
C=Aε/L

What I'm able to do so far :
The magnitude of the charge on the plates as a function of length L and ε should be
Q=CV_0=A_0ε/L × V_0 = A_0 V_0 ε / L
So the force on the top plate should be
F_elec = QE = A_0 V_0² ε / 2 L²
So for the process to proceed quasi statically, the force due to pressure should be equal to F_elec
P A_0 = A_0 V_0² ε / 2 L²
P = V_0² ε / 2 L²
Now since V=A_0 L
P = A_0² V_0² k / 2 V² T
So we're getting
P V² T = constant
Since T = P V / n R ,
P² V³ = constant
So a = 1.5
So since L_final = 2 L_0
V_f = 2 V_0
As per the process T_final = T_0 / √2
Now ½ C V_0²
with C= A_0 ε / L should be the Helmholtz Free Energy
So the internal electromagnetic energy should be
U=F-T(dF/dT)
=½ A_0 V_0² / L [ ε - T dε/dT ]
= A_0 V_0² k / L T
So the total internal energy should be
U = ½ n f R T + A_0 V_0² k / L T
= 2 n R T + A_0 V_0² k / L T
So the net internal energy change should be
T : T_0 → T_0 / √2 , L : L_0 → 2 L_0
∆U = 2 n R T_0 [ 1/√2 -1 ] + A_0 V_0² k / L_0 T_0 [ 1/√2 -1 ]
Now if we use the initial equilibrium relation,
P_0 = V_0² k / 2 L_0² T_0
V_0 = A_0 L_0
P_0 V_0 = A_0 V_0 ² k / 2 L_0 T_0
P_0 V_0 = n R T_0
So we finally get
∆U = 4 n R T_0 [ 1/√2 -1 ]
Now to find ∆Q, taking the capacitor plates + the gas as our system,
∆U=∆Q+W_battery
W_battery = V_0 ∆q = V_0² ∆C
Upon calculating we again get
W_battery = 2 n R T_0 [ 1/√2 - 1 ]
So ∆Q= 2 n R T_0 [ 1/√2 -1 ]
Is this correct ? As I am not confident about the calculation of the internal energy of the dielectric.
In some places the mentioned expression for energy density is given as
u=½ E² [ ε + T dε/dT ]
But in this case I'm getting a negative sign.
Thanks in advance.

 

[Andrew Mason]

Homework Statement: Consider a long cylindrical vessel of base area A_0. The ends of the vessel are fitted with two massless metallic plates also of area A_0. The bottom plate is fixed while the top plate is free to move. The initial length of the cylindrical column between the plates is L_0, in which n moles of an ideal gas (f=4) are filled, initially at a temperature T_0. It has a hypothetical permittivity ε(T)=k/T, and acts as an isotropic dielectric. The walls of the vessel are electrically insulating but allow heat exchange. The two metallic ends are connected to the terminals of an ideal battery of constant emf V_0. Initially, the upper plate is in equilibrium. The gas is then subjected to a polytropic process such that the process is quasi static, so as to expand until the length of the column is 2L_0. Thus, find the polytropic index a of the process, and also the heat exchange (if any).
Relevant Equations: PV=nRT
dU=TdS-dW
P_elec=σ²/2ε
C=Aε/L

What I'm able to do so far :
The magnitude of the charge on the plates as a function of length L and ε should be
Q=CV_0=A_0ε/L × V_0 = A_0 V_0 ε / L

So the force on the top plate should be
F_elec = QE = A_0 V_0² ε / 2 L²

The use of capital V for volume and voltage is confusing. I will use lower case $v$ for voltage and upper case $V$ for volume.

Your result for $a=1.5$ looks right. Since $PV^a=\text{constant}=P_0V_0^a$ and $P=nRT/V$ then

$nRTV^{a-1}=nRT_0V_0^{a-1}$ so:

$\frac{T}{T_0}=\left(\frac{V}{V_0}\right)^{1-a}$ For $V=V_f=2V_0, T_f=2^{-.5}T_0$

So $\Delta U=nC_v\Delta T=nC_v(T_f-T_0)$ where $C_v=2R$ since there are 4 degrees of freedom.

This results in:
(1) $\Delta U=2nRT_0(2^{-.5}-1)=-(2-\sqrt{2})nRT_0$

Compare that to the work done by the gas: $W=\int_{L_0}^{2L_0} FdL=\int_{L_0}^{2L_0} (A_0 v_{0}^2k /2 L^2T)dL$
You have to substitute $T=T_0\left(\frac{V}{V_0}\right)^{1-a}=T_0\left(\frac{L}{L_0}\right)^{-.5}$

Solve that to find W. Since $\Delta U=Q-W$, $W$, the work done by the gas plus the (negative) change in internal energy, $\Delta U$, in (1)is the heat flow, $Q$.

AM

 

[frmn]

The use of capital V for volume and voltage is confusing. I will use lower case $v$ for voltage and upper case $V$ for volume.

Your result for $a=1.5$ looks right. Since $PV^a=\text{constant}=P_0V_0^a$ and $P=nRT/V$ then

$nRTV^{a-1}=nRT_0V_0^{a-1}$ so:

$\frac{T}{T_0}=\left(\frac{V}{V_0}\right)^{1-a}$ For $V=V_f=2V_0, T_f=2^{-.5}T_0$

So $\Delta U=nC_v\Delta T=nC_v(T_f-T_0)$ where $C_v=2R$ since there are 4 degrees of freedom.

This results in:
(1) $\Delta U=2nRT_0(2^{-.5}-1)=-(2-\sqrt{2})nRT_0$

Compare that to the work done by the gas: $W=\int_{L_0}^{2L_0} FdL=\int_{L_0}^{2L_0} (A_0 v_{0}^2k /2 L^2T)dL$
You have to substitute $T=T_0\left(\frac{V}{V_0}\right)^{1-a}=T_0\left(\frac{L}{L_0}\right)^{-.5}$

Solve that to find W. Since $\Delta U=Q-W$, $W$, the work done by the gas plus the (negative) change in internal energy, $\Delta U$, in (1)is the heat flow, $Q$.

AM

Firstly, thanks for the response ! And apologies for the use of $V$ for both the variables and lack of LaTeX, as this was posted in a hurry. One query I had, so we do not need to count in the capacitance energy as part of $U$ ? That was the largest source of confusion for me as the energy would presumably be stored in the polarization of gas particles, and the fact that $a=1.5$ matches the adiabatic index of the gas as $\gamma=1+\frac{2}{f}=1.5$. Furthermore, does this suggest that the process may release heat from the electromagnetic energy rather than thermodynamic one ? Again, thanks in advance !

 

[Andrew Mason]

Firstly, thanks for the response ! And apologies for the use of $V$ for both the variables and lack of LaTeX, as this was posted in a hurry. One query I had, so we do not need to count in the capacitance energy as part of $U$ ? That was the largest source of confusion for me as the energy would presumably be stored in the polarization of gas particles, and the fact that $a=1.5$ matches the adiabatic index of the gas as $\gamma=1+\frac{2}{f}=1.5$. Furthermore, does this suggest that the process may release heat from the electromagnetic energy rather than thermodynamic one ? Again, thanks in advance !

You have found an easier way to solve the problem by noticing that $a=3/2=\gamma$, the adiabatic constant (Cp=3R;Cv=2R). So the process is equivalent to an adiabatic process:->Q=0.

The work done by the gas in increasing the separation of the plates results is stored as electrical potential energy of the charges on the plates. We have to assume that the gas molecules are not affected by the electric field. So the change in capacitor energy does not affect the internal energy of the gas.

The main difficulty I have is in trying to figure out how the process can be made quasi-static.

AM

 

[frmn]

You have found an easier way to solve the problem by noticing that $a=3/2=\gamma$, the adiabatic constant (Cp=3R;Cv=2R). So the process is equivalent to an adiabatic process:->Q=0.

The work done by the gas in increasing the separation of the plates results is stored as electrical potential energy of the charges on the plates. We have to assume that the gas molecules are not affected by the electric field. So the change in capacitor energy does not affect the internal energy of the gas.

The main difficulty I have is in trying to figure out how the process can be made quasi-static.

AM

However, if we were to include the gas particle polarizations in the internal energy, $∆Q$ would clearly not be zero. Was my orginal attempt correct when assuming the gas particles are affected by the electric field ? As the problem does not explicitly mention to ignore the gas polarization... But assuming the gas particles are not affected and the gas only has a thermal internal energy, $∆Q=0$ . As for the quasi static nature, I hope so that this method is correct (?). For a quasi static process, the force on the plate should be extremely close to zero, thus the relation $P_{gas}A_0=F_{elec}$ should suffice to ensure it proceeds quasi statically, and it does end up giving a polytropic process with $a=1.5$. The effect on the gas particles should presumably not affect the process itself...