Work - expressions for ideal gases

We've already got some useful results for a variety of processes to find the work done in terms of pressures and volumes alone:

  • Isochoric process: $dv=0$, so $Pdv=0$, and
    $$\Delta w=0.$$
  • Isobaric process: $P$ is constant.
    $$\Delta w=P(V_2-V_1).$$
  • Isothermic process: for an ideal gas,
    $$\Delta w=RT\ln\left(\frac{v_2}{v_1}\right)=P_1 v_1 \ln\left(\frac{v_2}{v_1}\right).$$
  • New--to be derived today:

    Adiabatic process, ideal gas $$\Delta w=\frac{1}{1-\gamma}\left( P_2v_2-P_1v_1\right).$$

Now our goal is to find an expression for the work done by an ideal gas in an adiabatic (no heat transferred) process. This expression should depend only on the initial and final pressures and volumes, and not on $T$. (We'll need this later on when we calculate the efficiency of a Carnot heat engine.)

Adiabatic processes (ideal gas)

This is needed later in the course, for the calculation of the efficiency of a carnot cycle. See section 5.2 to make sure we're using the same form of final relationship with gamma.

Strategy: An adiabatic process is one in which $\delta q=0$. If we could find an expression for $P(v)$ (not involving $T$) we could integrate $\int P\,dv=\Delta w$ to get the work.

Outline

  1. We just showed two results in terms of internal energy and enthalpy which follow from the First Law: $$\delta q = dh-v\,dP; \ \ \delta q = du +P\,dV. \label{deltaq} $$
  2. For an ideal gas, enthalpy and internal energy depend on $T$ alone, which allows us write: $$\frac{dh}{dT}=c_P;\ \ \frac{du}{dT}=c_V.\label{dees}$$
  3. Using the expressions in Eq ($\ref{dees}$), substitute in for $dh$ and $du$ in Eq ($\ref{deltaq}$), and set $\delta q=0$ (adiabatic process).
  4. We need to get rid of temperature dependence: rearrange both expressions so that you have two equations with an expression involving $dT$ on one side. Now divide one equation by the other. You should have one equation that involves $P$, $dP$, $v$, $dv$ and $c_P$ and $c_V$.
  5. Gather all the terms involving $v$ and $dv$ on one side, and the terms with $P$ and $dP$ on the other side; For the ratio of heat capacities use this new constant, $\gamma$: $$\gamma \equiv \frac{c_P}{c_V}.$$
  6. Integrate both sides of your differential equation. This will give you your relationship $P(v)$.

One expression relating heat and enthalpy was... $$\delta q = dh-v\,dP.$$

For an ideal gas (only!), enthalpy depends *only* on $T$, that is $h=h(T)$, which implied $dh/dT=c_P$, so, this becomes... $$0=c_PdT-v\, dP \Rightarrow v\,dP=c_P dT.$$

Also, the internal energy, $u$, of an ideal gas depends only on $T$, and therefore $c_v=du/dT$. $$\begineq \delta q &= du +\delta w\\ 0&=c_vdT +P\,dv\endeq $$ So... $$\Rightarrow P\,dv=-c_v dT.$$ [Adiabatic process. Ideal gas only.]

Dividing... $$\frac{v\,dP}{P\,dv} = -\frac{c_P}{c_v}\equiv -\gamma.$$

...

Gathering similar terms: $$\frac{dP}{P}=-\gamma\frac{dv}{v}.$$

Integrating both sides: $$\ln(P) = -\gamma\ln(v) +C \Rightarrow \color{blue}{Pv^{\gamma}=e^C=K}\label{K}.$$

Solve this for $P(v)$, and integrate $\int P(v)dv=\Delta w$ to get the work done...

$$\begineq\Delta w = \int_{v_1}^{v_2} \frac{K}{v^\gamma}\,dv &= \int Kv^{-\gamma}\,dv\\ &= \left. \frac{Kv^{(-\gamma+1)}}{-\gamma+1} \right|_{v_1}^{v_2}\\ &= \left(\frac{1}{1-\gamma}\right)\left[ Kv_2^{(-\gamma+1)}-Kv_1^{(-\gamma+1)} \right] \endeq \label{PvStuff}$$

Because of Eq ($\ref{K}$), we could write $K$ in both of these ways, $$K=P_1v_1^\gamma=P_2v_2^\gamma$$ So, substituting these in for one $K$ or the other in Eq ($\ref{PvStuff}$)...

$$\Delta w=\frac{1}{1-\gamma}\left( P_2v_2-P_1v_1\right).$$

[Adiabatic process. Ideal gas.]

Referring back to this relation for an ideal gas, adiabatic process Eq ($\ref{K}$), we found that a product of $P$ and a power of $v$ is constant for an adiabatic process: $$Pv^{\gamma}=K.$$

Use the ideal gas law, and the relation above, to find a product of $T$ and a (different) power of $v$ which is constant.