Work done in adiabatic process

Question

When we try to establish a relation between the pressure and temperature in adiabatic process we come across a equation..

$dU = dq - PdV$

$dq=0$ (Adiabatic process) and,

$dU=C_v.dT$ (Heat capacity at constant volume)

Therefore, $C_v.dT = -PdV$$\tag1$

In this equation we are using heat capacity defined at constant volume but their still is some work done by the system (i.e, $PdV$ is not $0$ or $dV$ not equals to $0$).

The first part of the equation $(1)$ is implying that the volume is constant but the second part is implying that the volume is not constant (if it was there would be no work done).

Then why there is this contradiction ?

Answer 1

Even though we call $C_v$ the heat capacity at constant volume, what we really mean by the subscript v is that this is the way we measure $C_v$. At constant volume, we can determine the heat capacity of the material by measuring the heat transferred $dQ=dU=C_vdT$.

But this same heat capacity also applies to all other situations for an ideal gas if we recognize that, for an ideal gas, $U$ depends only on temperature, such that $dU=C_vdT$. It is just that, in these other situations (involving work), dQ is not equal to $C_vdT$.

Answer 2

I will just expand on what Chester Miller had already said.

The molar specific heat at constant volume $C_v$ is basically the amount of heat required to raise the temperature of 1 mol of the gas by $1$ K while keeping the volume constant.

When we apply heat to an ideal gas, some of that heat is used to increase the internal energy $U$ of the gas and the rest is eventually used up to do the work to increase its volume. (Note that increasing U essentially means increasing the temperature since U is a function of T for any ideal gas.) Now if we were to keep the volume constant somehow, then the total applied heat would be used only to increase the internal energy or so to speak the temperature and nothing else.

So a different way to define $C_v$ is to say, it is the amount of heat required to raise the temperature of 1 mol of the gas by $1$ K, where the said heat is used to only to change the internal energy U or so to speak the temperature and nothing else.

Now let's consider a case where change in volume is allowed. Say we applied $dQ$ amount of heat to the a gas. Now we know some of that heat is going to account for the change in internal energy, let's say that amount is $dQ_u$. And the rest of the heat will eventually account for the change in volume, and say that is $dQ_w$. So instead of writing $dQ = dU + dW$ we write, $$dQ = dQ_u +dQ_w$$

So we can see that the heat $dQ_u$ is only used to change the internal energy and not the volume and so $dQ_u = dU$ and we can say according to the definition of $C_v$ the following $$C_v = \frac{dQ_u}{mdT}\ \Rightarrow\ dQ_u = mC_vdT\ \Rightarrow\ dU=mC_vdT$$

So, you see $dU = mC_vdT$ whether or not there is any change in volume. The same argument goes for the adiabatic process you mentioned.

Answer 3

The answer is pretty simple. We will prove it for an ideal monoatomic gas, the proof can be adjusted for other gasses as well.

Firstly, I will use the convention that $\Delta E_{int} = Q + W$ where the sign of $Q$ is positive if heat is transferred to the system and the sign of $W$ is positive if the positive work is done to the system (by the environment).

Note that we have $E_{int} = \frac{3}{2}nRT$ for a monoatomic gas, and thus $\Delta E_{int} = \frac{3}{2} nR\Delta T.$ What this essentially means is that regardless of the process, the change of the internal energy depends only on the final and initial temperature.

If the volume of the gas is constant, then $W = 0$, so $Q = \Delta E_{int}$. By the definition of $C_v$, we have $$C_v = \frac{Q}{n\Delta T} =\frac{\Delta E_{int}}{n\Delta T}$$ $$\implies \Delta E_{int} = C_v n \Delta T.$$ Now equating the two equations give $C_v = \frac{3}{2}R$.

Note that $\Delta E_{int} = \frac{3}{2} nR\Delta T$ applies in any process, and thus $\Delta E_{int} = nC_v dT$ also applies in any process since they are the SAME equation, and that $C_v$ is independent of the type of the monoatomic gas.

Answer 4

Thermal capacity at constant volume $\textrm C_V$ is defined as

$$\mathrm C_V ~=~\left(\frac{\partial U}{\partial T}\right)_V$$ where $U$ is the internal energy of the system.

For an ideal gas, $U=U(T);$ so, $$\mathrm C_V~\mathrm dT ~=~ \mathrm dU\,.\tag I$$

Substituting $\mathrm{(I)}$ in the First Law of Thermodynamics,

$$\mathrm C_V~\mathrm dT +đw~=~ đQ\tag{II} $$

From which, for an adiabatic process, $$\mathrm C_V~\mathrm dT ~=-~P~\mathrm dV,$$

which is actually $$\mathrm dU ~= -~P~\mathrm dV\,.$$

Nothing is contradictory here.