3.6 Adiabatic Processes for an Ideal Gas - University Physics Volume 2

Learning Objectives

By the end of this section, you will be able to:

Define adiabatic expansion of an ideal gas
Demonstrate the qualitative difference between adiabatic and isothermal expansions

When an ideal gas is compressed adiabatically $(Q = 0),$ work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops. Adiabatic compressions actually occur in the cylinders of a car, where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heat with its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does rise significantly. In fact, the temperature increases can be so large that the mixture can explode without the addition of a spark. Such explosions, since they are not timed, make a car run poorly—it usually “knocks.” Because ignition temperature rises with the octane of gasoline, one way to overcome this problem is to use a higher-octane gasoline.

Another interesting adiabatic process is the free expansion of a gas. Figure 3.13 shows a gas confined by a membrane to one side of a two-compartment, thermally insulated container. When the membrane is punctured, gas rushes into the empty side of the container, thereby expanding freely. Because the gas expands “against a vacuum” $(p = 0)$ , it does no work, and because the vessel is thermally insulated, the expansion is adiabatic. With $Q = 0$ and $W = 0$ in the first law, $Δ E_{int} = 0,$ so $E_{int}_{i} = E_{int}_{f}$ for the free expansion.

The figure on the left is an illustration of the initial equilibrium state of a container with a partition in the middle dividing it into two chambers. The outer walls are insulated. The chamber on the left is full of gas, indicated by blue shading and many small dots representing the gas molecules. The right chamber is empty. The figure on the right is an illustration of the final equilibrium state of the container. The partition has a hole in it. The entire container, on both sides of the partition, is full of gas, indicated by blue shading and many small dots representing the gas molecules. The dots in the second, final equilibrium state, illustration are less dense than in the first, initial state illustration.

Figure 3.13 The gas in the left chamber expands freely into the right chamber when the membrane is punctured.

If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, its temperature does not change.

A quasi-static, adiabatic expansion of an ideal gas is represented in Figure 3.14, which shows an insulated cylinder that contains 1 mol of an ideal gas. The gas is made to expand quasi-statically by removing one grain of sand at a time from the top of the piston. When the gas expands by dV, the change in its temperature is dT. The work done by the gas in the expansion is $d W = p d V; d Q = 0$ because the cylinder is insulated; and the change in the internal energy of the gas is, from Equation 3.9, $d E_{int} = C_{V} n d T .$ Therefore, from the first law,

C_{V} n d T = 0 - p d V = − p d V,

d T = - \frac{p d V}{C_{V} n} .

The figure is an illustration of a container. The walls and bottom are filled with a thick layer of insulation. The chamber of the container is closed from above by a piston. Inside the chamber is a gas. There is a pile of sand on top of the piston, and a hand with tweezers is removing grains from the pile.

Figure 3.14 When sand is removed from the piston one grain at a time, the gas expands adiabatically and quasi-statically in the insulated vessel.

Also, for 1 mol of an ideal gas,

d (p V) = d (R n T),

p d V + V d p = R n d T

and

d T = \frac{p d V + V d p}{R n} .

We now have two equations for dT. Upon equating them, we find that

C_{V} n V d p + (C_{V} n + R n) p d V = 0 .

Now, we divide this equation by npV and use $C_{p} = C_{V} + R$ . We are then left with

C_{V} \frac{d p}{p} + C_{p} \frac{d V}{V} = 0,

which becomes

\frac{d p}{p} + γ \frac{d V}{V} = 0,

where we define $γ$ as the ratio of the molar heat capacities:

γ = \frac{C_{p}}{C_{V}} .

3.11

Thus,

\int \frac{d p}{p} + γ \int \frac{d V}{V} = 0

and

ln p + γ ln V = constant .

Finally, using $ln (A^{x}) = x ln A and ln A B = ln A + ln B$ , we can write this in the form

p V^{γ} = constant .

3.12

This equation is the condition that must be obeyed by an ideal gas in a quasi-static adiabatic process. For example, if an ideal gas makes a quasi-static adiabatic transition from a state with pressure and volume $p_{1}$ and $V_{1}$ to a state with $p_{2}$ and $V_{2},$ then it must be true that $p_{1} V_{1}^{γ} = p_{2} V_{2}^{γ} .$

The adiabatic condition of Equation 3.12 can be written in terms of other pairs of thermodynamic variables by combining it with the ideal gas law. In doing this, we find that

p^{1 - γ} T^{γ} = constant

3.13

and

T V^{γ - 1} = constant .

3.14

A reversible adiabatic expansion of an ideal gas is represented on the pV diagram of Figure 3.15. The slope of the curve at any point is

\frac{d p}{d V} = \frac{d}{d V} (\frac{constant}{V^{γ}}) = − γ \frac{p}{V} .

The figure is a plot of pressure, p on the vertical axis as a function of volume, V on the horizontal axis. Two curves are plotted. Both are monotonically decreasing and concave up. One is slightly higher and has a greater curvature. This curve is labeled “isothermal.” The second curve is below the isothermal curve and has a slightly smaller curvature. This curve is labeled “adiabatic.”

Figure 3.15 Quasi-static adiabatic and isothermal expansions of an ideal gas.

The dashed curve shown on this pV diagram represents an isothermal expansion where T (and therefore pV) is constant. The slope of this curve is useful when we consider the second law of thermodynamics in the next chapter. This slope is

\frac{d p}{d V} = \frac{d}{d V} \frac{n R T}{V} = - \frac{p}{V} .

Because $γ > 1,$ the isothermal curve is not as steep as that for the adiabatic expansion.

Example 3.7

Compression of an Ideal Gas in an Automobile Engine

Gasoline vapor is injected into the cylinder of an automobile engine when the piston is in its expanded position. The temperature, pressure, and volume of the resulting gas-air mixture are

20 ° C

1.00 \times 10^{5} {N/m}^{2},

and

240 {cm}^{3}

, respectively. The mixture is then compressed adiabatically to a volume of

40 {cm}^{3}

. Note that in the actual operation of an automobile engine, the compression is not quasi-static, although we are making that assumption here. A second, important approximation is that although the vapor is a gasoline-air mixture of nitrogen, oxygen, carbon dioxide, water, and hydrocarbons, we treat it as an ideal, diatomic gas. (a) What are the pressure and temperature of the mixture after the compression? (b) How much work is done by the mixture during the compression?

Strategy

Because we are modeling the process as a quasi-static adiabatic compression of an ideal gas, we have

p V^{γ} = constant

and

p V = n R T

. The work needed can then be evaluated with

W = \int_{V_{1}}^{V_{2}} p d V

Solution

For an adiabatic compression we have
$p_{2} = p_{1} {(\frac{V_{1}}{V_{2}})}^{γ} .$
Since we are treating the vapor as an ideal, diatomic gas, we can use $γ = \frac{7}{5} = 1.4$ . So after the compression, the pressure of the mixture is
$p_{2} = (1.00 \times 10^{5} {N/m}^{2}) {(\frac{240 \times 10^{−6} m^{3}}{40 \times 10^{−6} m^{3}})}^{1.40} = 1.23 \times 10^{6} {N/m}^{2} .$
From the ideal gas law, the temperature of the mixture after the compression is
$\begin{matrix} T_{2} & = (\frac{p_{2} V_{2}}{p_{1} V_{1}}) T_{1} \\ = \frac{(1.23 \times 10^{6} {N/m}^{2}) (40 \times 10^{−6} m^{3})}{(1.00 \times 10^{5} {N/m}^{2}) (240 \times 10^{−6} m^{3})} \cdot 293 K \\ = 600 K = 328 ° C . \end{matrix}$
The work done by the mixture during the compression is
$W = \int_{V_{1}}^{V_{2}} p d V .$
With the adiabatic condition of Equation 3.12, we may write p as $K / V^{γ},$ where $K = p_{1} V_{1}^{γ} = p_{2} V_{2}^{γ} .$ The work is therefore
$\begin{matrix} W & = \int_{V_{1}}^{V_{2}} \frac{K}{V^{γ}} d V \\ = \frac{K}{1 - γ} (\frac{1}{V_{2}^{γ - 1}} - \frac{1}{V_{1}^{γ - 1}}) \\ = \frac{1}{1 - γ} (\frac{p_{2} V_{2}^{γ}}{V_{2}^{γ - 1}} - \frac{p_{1} V_{1}^{γ}}{V_{1}^{γ - 1}}) \\ = \frac{1}{1 - γ} (p_{2} V_{2} - p_{1} V_{1}) \\ = \frac{1}{1 - 1.40} [(1.23 \times 10^{6} {N/m}^{2}) (40 \times 10^{−6} m^{3}) \\ - (1.00 \times 10^{5} {N/m}^{2}) (240 \times 10^{−6} m^{3})] \\ = −63 J . \end{matrix}$

Significance

The negative sign on the work done indicates that the piston does work on the gas-air mixture. The engine would not work if the gas-air mixture did work on the piston.