4.4 - Gas Laws

Pressure, Temperature and Volume

Gas molecules move randomly and freely throughout their container.
Pressure, $P$ : molecules collide with the container walls and rebound, exerting a force; force per unit area gives gas pressure. S.I. unit: pascal, Pa; another pressure unit is cm Hg.
Temperature: higher temperature means greater average molecular kinetic energy. Absolute temperature, $T$ , is measured in kelvin, K; other temperature units include $^{\circ} C$ and $^{\circ} F$ .
Volume, $V$ : gas fills the available space, so its volume equals the container’s volume. S.I. unit: $m^{3}$ ; other units include $c m^{3}$ , $m m^{3}$ and ml.

Inference: volume influences gas pressure.
Aim: determine the pressure–volume relationship for a fixed mass of gas at constant temperature.
Hypothesis: smaller volume gives higher pressure.
Variables: manipulated — volume $V$ ; responding — pressure $P$ ; constant — temperature and mass of air.
Apparatus: 100 ml syringe, rubber tube, pressure gauge and retort stand.

Set the syringe to 100 ml, then connect its outlet to the pressure gauge using the rubber tube.
Record initial volume and pressure.
Push the piston slowly to 90 ml and record pressure; repeat for 80, 70 and 60 ml.
Tabulate $V$ (ml), $P$ (kPa) and $1/ V$ ( $m l^{- 1}$ ); plot $P$ against $V$ and $P$ against $1/ V$ .

Result and conclusion: pressure rises as volume decreases; the straight $P$ – $1/ V$ graph through the origin establishes $P \propto 1/ V$ .

For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume:

P \propto \frac{1}{V}, P V = k, P_{1} V_{1} = P_{2} V_{2}

Graphs: $P$ against $V$ is a decreasing curve; $P$ against $1/ V$ is a straight line through the origin.
Molecular explanation: reducing volume puts the same number of molecules into a smaller space. Molecules per unit volume increase, wall collisions occur more frequently, and force per unit area increases.
Example: 60 $c m^{3}$ of air at 108 kPa is compressed to 48 $c m^{3}$ without a temperature change:

P_{2} = \frac{108 ( 60 )}{48} = 135 kPa

Inference: temperature influences gas volume.
Aim: determine the volume–temperature relationship for a fixed mass of gas at constant pressure.
Hypothesis: higher temperature gives greater volume.
Variables: manipulated — temperature $θ$ ; responding — volume $V$ , represented by air-column length $L$ ; constant — pressure and mass of air.
Apparatus: capillary tube with air trapped by concentrated sulphuric acid, 500 ml beaker, thermometer, ruler, Bunsen burner, tripod stand, wire gauze, stirrer and retort stand. Materials: water, ice and rubber band.

Place the capillary tube, ruler and thermometer in the water bath, with the trapped air immersed.
Heat slowly and stir continuously until the water reaches $30^{\circ} C$ ; record $L$ .
Repeat at 40, 50, 60, 70 and $80^{\circ} C$ , tabulating temperature and $L$ (cm).
Plot $L$ against $θ$ with an axis from 0 to $100^{\circ} C$ ; extrapolate to $0^{\circ} C$ .
Replot with an axis from $- 300$ to $100^{\circ} C$ and extrapolate to $L = 0$ .

The volume–length relationship is $V = A L$ , where $A$ is the capillary’s uniform cross-sectional area, so $L$ represents volume.
Result and conclusion: volume rises with temperature; its non-zero value at $0^{\circ} C$ shows that molecules still move. Extrapolation gives absolute zero, and the kelvin graph establishes $V \propto T$ .

At $0^{\circ} C$ , gas molecules still move; the gas has volume and pressure.
Extrapolating the volume–Celsius-temperature graph at constant pressure to zero volume, or the pressure–Celsius-temperature graph at constant volume to zero pressure, gives $- 273^{\circ} C$ .
This is absolute zero, the lowest possible temperature, corresponding to 0 K. Temperature expressed in kelvin is absolute temperature.

T = θ + 273

$θ$ is the numerical temperature in $^{\circ} C$ ; $T$ is in K. Melting ice: 273 K; steam at $100^{\circ} C$ : 373 K.

For a fixed mass of gas at constant pressure, volume is directly proportional to absolute temperature:

V \propto T, \frac{V}{T} = k, \frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}

Graphs: $V$ against Celsius temperature increases linearly and extrapolates to zero at $- 273^{\circ} C$ ; $V$ against $T$ is a straight line through the origin.
Molecular explanation: heating increases average kinetic energy and molecular velocity. Volume expands to maintain constant pressure; the wall-collision rate remains unchanged.
Example: a 1.2 $c m^{3}$ air bubble warms from 27 to $47^{\circ} C$ at constant pressure:

T_{1} = 300 K, T_{2} = 320 K, V_{2} = \frac{1.2 ( 320 )}{300} = 1.28 c m^{3}

Inference: temperature influences gas pressure.
Aim: determine the pressure–temperature relationship for a fixed mass of gas at constant volume.
Hypothesis: higher temperature gives higher pressure.
Variables: manipulated — temperature $θ$ ; responding — pressure $P$ ; constant — volume and mass of air.
Apparatus: round-bottom flask, large beaker, thermometer, pressure gauge, Bunsen burner, wire gauze, tripod stand, stirrer, retort stand and wooden block. Materials: water and ice.

Immerse the flask containing trapped air in a water bath; connect it to the pressure gauge with a rubber tube. Place the thermometer and stirrer in the bath.
Heat slowly and stir continuously until the bath reaches $30^{\circ} C$ ; record air pressure.
Repeat at 40, 50, 60, 70 and $80^{\circ} C$ , tabulating temperature and pressure (kPa).
Plot $P$ against $θ$ with an axis from $- 300$ to $100^{\circ} C$ ; extrapolate to $P = 0$ .

Result and conclusion: pressure rises with temperature; extrapolation gives $- 273^{\circ} C$ , while the kelvin graph establishes $P \propto T$ .

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature:

P \propto T, \frac{P}{T} = k, \frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}

Graphs: $P$ against Celsius temperature increases linearly and extrapolates to zero at $- 273^{\circ} C$ ; $P$ against $T$ is a straight line through the origin.
Molecular explanation: heating increases average kinetic energy and molecular velocity. At unchanged volume, molecules collide with walls more frequently, increasing force per unit area and pressure.
Example: gas at 180 kPa in a closed steel cylinder warms from 25 to $52^{\circ} C$ at constant volume:

T_{1} = 298 K, T_{2} = 325 K, P_{2} = \frac{180 ( 325 )}{298} = 196.3 kPa

Identify the constant quantity and retain fixed gas mass. Use consistent pressure units and consistent volume units; convert temperatures to K for temperature ratios.
Constant temperature: a syringe changes from 7.5 to 2.5 $c m^{3}$ , initially at 105 kPa:

P_{2} = \frac{105 ( 7.5 )}{2.5} = 315 kPa

V_{2} = \frac{0.24 ( 350 )}{300} = 0.28 m^{3}

Constant volume: tyre pressure rises from 210 to 240 kPa, initially at $25^{\circ} C$ . Assume the tyre volume does not change:

T_{2} = \frac{240 ( 298 )}{210} = 340.6 K, θ_{2} = 340.6 - 273 = 67.6^{\circ} C