18.7.4 Constant-Pressure Process

Other gas processes take place at a constant, unchanging pressure. A constant-pressure process is called an isobaric process, where baric is from the same root as “barometer” and means “pressure.” An isobaric process is one for which

p_{f} = p_{i}

(18.20)

FIGURE 18.12a shows one method of changing the state of a gas while keeping the pressure constant. A cylinder of gas has a tight-fitting piston of mass $M$ that can slide up and down but seals the container so that no atoms enter or escape. As the free-body diagram of FIGURE 18.12b shows, the piston and the air press down with force $p_{atmos} A + M g$ while the gas inside pushes up with force $p_{gas} A$ . In equilibrium, the gas pressure inside the cylinder is

p_{gas} = p_{atmos} + \frac{M g}{A}

(18.21)

In other words, the gas pressure is determined by the requirement that the gas must support both the mass of the piston and the air pressing inward. This pressure is independent of the temperature of the gas or the height of the piston, so it stays constant as long as $M$ is unchanged.

The figure labeled ("a") consists of two parts labeled before and after that illustrate the change in pressure at constant volume.

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The figure labeled (b) illustrates a free body diagram for part ("a").

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The figure labeled (c) illustrates a pressure versus volume graph.

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If the cylinder is warmed, the gas will expand and push the piston up. But the pressure, determined by mass $M$ , will not change. This process is shown on the $p V$ diagram of FIGURE 18.12c as the horizontal line $1 \to 2$ . We call this an isobaric expansion. An isobaric compression occurs if the gas is cooled, lowering the piston. Any isobaric process appears on a $p V$ diagram as a horizontal line.

EXAMPLE 18.8 Comparing pressure

The two cylinders in FIGURE 18.13 contain ideal gases at $20 ° C$ . Each cylinder is sealed by a frictionless piston of mass $M$ .

How does the pressure of gas 2 compare to that of gas 1? Is it larger, smaller, or the same?
Suppose gas 2 is warmed to $80 ° C$ . Describe what happens to the pressure and volume.

The figure consists of two parts labeled Gas 1 and Gas 2 that illustrate two gases inside a piston at 20 degrees Celsius.

FIGURE 18.13 Full Alternative Text

MODEL Treat the gases as ideal gases.

SOLVE

The pressure in the gas is determined by the requirement that the piston be in mechanical equilibrium. The pressure of the gas inside pushes up on the piston; the air pressure and the weight of the piston press down. The gas pressure $p = p_{atmos} + M g / A$ depends on the mass of the piston, but not at all on how high the piston is or what type of gas is inside the cylinder. Thus both pressures are the same.
Neither does the pressure depend on temperature. Warming the gas increases the temperature, but the pressure—determined by the mass and area of the piston—is unchanged. Because $p V / T = constant,$ and $p$ is constant, it must be true that $V / T = constant$ . As $T$ increases, the volume $V$ also must increase to keep $V / T$ unchanged. In other words, increasing the gas temperature causes the volume to expand—the piston goes up—but with no change in pressure. This is an isobaric process.

EXAMPLE 18.9 Identifying a gas

Your lab assistant distilled 50 g of a gas into a cylinder, but he left without writing down what kind of gas it is. The cylinder has a pressure regulator that adjusts a piston to keep the pressure at a constant 2.00 atm. To identify the gas, you measure the cylinder volume at several different temperatures, acquiring the data shown at the right. What is the gas?

$T (° C)$	$V (L)$
$- 50$	11.6
0	14.0
50	16.2
100	19.4
150	21.8

MODEL The pressure doesn’t change, so heating the gas is an isobaric process.

SOLVE The ideal-gas law is $p V = n R T$ . Writing this as

V = \frac{n R}{p} T

we see that a graph of $V$ versus $T$ should be a straight line passing through the origin. Further, we can use the slope of the graph, $n R / p$ , to measure the number of moles of gas, and from that we can identify the gas by determining its molar mass.

FIGURE 18.14 is a graph of the data, with the volumes and temperatures converted to SI units of $m^{3}$ $(1 m^{3} = 1000 L)$ and kelvins. The $y$ -intercept of the graph is essentially zero, confirming the behavior of the gas as ideal, and the slope of the best-fit line is $5.16 \times 10^{- 5} m^{3} / K$ . The number of moles of gas is

n = \frac{p}{R} \times slope = \frac{2 \times 101, 300 Pa}{8.31 J / mol K} \times 5.16 \times 10^{- 5} m^{3} / K = 1.26 mol

The figure illustrates a gas volume versus temperature graph.

FIGURE 18.14 Full Alternative Text

From this, the molar mass is

M = \frac{0.050 kg}{1.26 mol} = 0.040 kg / mol

Thus the atomic mass is 40 u, identifying the gas as argon.

REVIEW The atomic mass is that of a well-known gas, which gives us confidence in the result.

18.7.4 Constant-Pressure Process

FIGURE 18.12 A constant-pressure (isobaric) process.

EXAMPLE 18.8 Comparing pressure

FIGURE 18.13 Compare the pressures of the two gases.

EXAMPLE 18.9 Identifying a gas

FIGURE 18.14 A graph of the gas volume versus its temperature.