Lesson 9.3 · Unit 9 · CED topic 9.2
Gas-law graphs and experiments
In the 1600s Robert Boyle trapped air in a bent glass tube, added mercury to squeeze it, and measured how the volume shrank as the pressure grew. Two centuries later experimenters heated gases and noticed something odd: when they extended their graphs of pressure against temperature backward, every gas pointed to the same temperature, about \(-273\,^\circ\)C. That unexpected agreement was the first hint of an absolute zero. This lesson turns the ideal gas law into graphs you can sketch and read, shows how to extract physical constants from them, and describes how you would test the law in a laboratory.
Describe the properties of an ideal gas, including how graphs of its pressure, volume and temperature are used.
- Graphs modeling the pressure, temperature, and volume of gases can be used to describe or determine properties of that gas.
- A temperature at which an ideal gas has zero pressure can be extrapolated from a graph of pressure as a function of temperature.
Four graphs, one law
Start from \(PV = nRT\) and hold two quantities fixed. Each choice gives a graph with a different shape, and the skill is to recognize each shape and to know what its slope means.
| Fixed | Graph | Shape | Slope or constant |
|---|---|---|---|
| \(n\), \(T\) | \(P\) against \(V\) | Hyperbola (an isotherm) | \(PV = nRT\) at every point |
| \(n\), \(T\) | \(P\) against \(1/V\) | Straight line through the origin | Slope \(= nRT\) |
| \(n\), \(V\) | \(P\) against \(T\) (kelvin) | Straight line through the origin | Slope \(= nR/V\) |
| \(n\), \(P\) | \(V\) against \(T\) (kelvin) | Straight line through the origin | Slope \(= nR/P\) |
The first row is the one to learn well. The graph of \(P\) against \(V\) curves, so you cannot read a constant from it directly. Plot \(P\) against \(1/V\) instead and the data fall on a straight line. This is linearization: choose the variables so that the relationship becomes a straight line and the slope carries the physics. In a real experiment you do this because a line is easy to fit and easy to judge by eye, while a curve is not.
Changing the temperature changes the graph in predictable ways. On a \(P\)–\(V\) graph a higher temperature gives an isotherm farther from the origin (a larger \(PV\) product). On a \(P\)–\(1/V\) graph a higher temperature gives a steeper line. If you are asked to sketch two isotherms, draw the one for the higher temperature above and to the right of the other, and never let them cross.
Absolute zero from a graph
Fix the amount of gas and the volume, and measure the pressure at several Celsius temperatures. The points fall on a straight line, but the line does not pass through the origin of the Celsius axis, because 0\(^\circ\)C is not a special temperature for a gas. Extend the line backward until it reaches \(P = 0\). The temperature where this happens is the extrapolated absolute zero, near \(-273\,^\circ\)C. In kelvin the line passes through the origin, consistent with \(P = (nR/V)T\).
Notice that this is an extrapolation. Real gases condense to liquids before they get anywhere near this temperature, so no real gas actually reaches zero pressure. What the graph shows is that the ideal gas model, extended, points to a lowest possible temperature. Every gas gives nearly the same intercept, which is evidence that the temperature scale from the gas law is universal.
Designing a gas-law experiment
The free-response section often asks you to design an experiment. For gases the classic design is a sealed syringe. Trap a fixed amount of air in a syringe, attach a pressure sensor, and change the volume slowly (so the temperature stays constant). A second version loads books of known weight on the plunger: the absolute pressure is \(P = P_{atm} + mg/A\), where \(A\) is the cross-sectional area of the plunger.
- Measure. Record the volume \(V\) from the syringe scale and the pressure \(P\) from the sensor for at least five settings.
- Control. Move the plunger slowly so the air stays at room temperature, and check for leaks, since \(n\) must stay fixed.
- Plot. Graph \(P\) against \(1/V\). The data should fall on a line through the origin with slope \(nRT\).
- Analyze. Find the slope of the best-fit line. Then \(n = \text{slope}/(RT)\), or compare the slope with the prediction.
- Error sources. Air trapped in the tip of the syringe adds a small volume that does not change. Friction between the plunger and the barrel changes the pressure for the books-on-plunger method. Slow changes only keep the temperature constant approximately.
A student measures the pressure of a fixed volume of gas: 100 kPa at 0\(^\circ\)C and 137 kPa at 100\(^\circ\)C. Assume the pressure is a linear function of temperature and extrapolate to the temperature at which the pressure would be zero.
Slope. \((137-100)\ \text{kPa}/(100\,^\circ\text{C}) = 0.37\) kPa/\(^\circ\)C.
Extrapolate. From 100 kPa at 0\(^\circ\)C, the pressure falls to zero after \(100/0.37 = 270\) degrees below zero, so the intercept is \(-270\,^\circ\)C.
Compare. The accepted value is \(-273.15\,^\circ\)C; the difference of about 1% is within the precision of two data points.
A gas in a syringe is at 100 kPa and 2.0 mL. The plunger is pushed slowly until the volume is 0.50 mL, and the temperature stays constant. Find the final pressure, and state what a \(P\)–\(1/V\) graph would show.
Isotherm. \(P_1V_1 = P_2V_2\), so \(P_2 = (100)(2.0)/0.50 = 400\) kPa.
Graph. The factor of 4 in \(1/V\) gives the factor of 4 in \(P\). The points lie on a straight line through the origin with slope \(P_1V_1 = 200\) kPa·mL.
A sealed 2.0 L container of air at 300 K and \(1.0\times10^5\) Pa is heated to 450 K while a piston lets the volume grow to 3.0 L. Find the final pressure.
Equation. \(P_2 = P_1(V_1/V_2)(T_2/T_1)\).
Substitute. \(P_2 = (1.0\times10^5)(2.0/3.0)(450/300) = 1.0\times10^5\) Pa. The pressure is unchanged: the volume grew by a factor of 1.5 and so did the temperature.
- Shapes. \(P\)–\(V\) hyperbola; \(P\)–\(1/V\), \(P\)–\(T\) and \(V\)–\(T\) straight lines through the origin (kelvin).
- Slopes. \(P\) against \(1/V\): \(nRT\). \(P\) against \(T\): \(nR/V\). \(V\) against \(T\): \(nR/P\).
- Absolute zero. Extrapolate the \(P\)–\(T\) line back to \(P = 0\); it is about \(-273\,^\circ\)C for any gas.
- Experiment design. Measure \(P\) and \(V\); keep \(T\) and \(n\) constant; linearize with \(1/V\); get the constant from the slope.
- Sketching. Higher temperature: isotherm farther from the origin, steeper \(P\)–\(1/V\) line. Isotherms never cross.
- A gas at fixed temperature has a pressure of 60 kPa at 5.0 L. What is its pressure at 3.0 L?
Show answer
\(P_2 = (60)(5.0)/3.0 = 100\) kPa.
- On a graph of \(P\) against \(1/V\) for a fixed amount of gas at 300 K, the slope is 24.9 J (in SI units). How many moles are there?
Show answer
Slope \(= nRT\), so \(n = 24.9/(8.31\times300) = 0.010\) mol.
- A student plots the pressure of a fixed gas sample against its Celsius temperature and gets a straight line that reaches zero pressure at \(-272\,^\circ\)C. What does the intercept represent, and why is it an extrapolation?
Show answer
It estimates absolute zero, the temperature at which the ideal gas pressure would vanish. It is an extrapolation because real gases condense before reaching it, so no measurement can be made there.
- A student puts books on a syringe plunger and notices that the air volume is smaller than the ideal gas law predicts at the first setting. Name one experimental error that could explain this.
Show answer
Friction in the plunger changes the net force that the books apply, a small leak lets gas escape (smaller \(n\)), or air was heated by the pushing; any of these changes \(P\) or \(n\) from what was assumed.
- Sketch two isotherms on a \(P\)–\(V\) graph, one at 300 K and one at 600 K. Which one is farther from the origin?
Show answer
The 600 K isotherm is farther from the origin because \(PV = nRT\) is twice as large. The curves are hyperbolas that do not cross.
- A fixed amount of gas in a rigid container is heated. A student plots \(P\) against \(T\) in kelvin. What does the graph look like, and what physical quantities does the slope contain?
Show answer
A straight line through the origin. The slope is \(nR/V\) (or \(Nk_B/V\)), which depends on the amount of gas and the container volume.