07/14 Gases and Molecular Motion

A progressive guide to gas properties, gas laws, molecular motion, mixtures, reaction calculations, and experimental reliability.

The Four Variables of Gas Behavior

Gas behavior is controlled mainly by four variables: pressure, volume, amount, and temperature. A gas has no fixed shape or volume, expands to fill its container, has low density because its particles are widely separated, and is highly compressible. Gas particles also diffuse through space and effuse through very small openings.

Pressure is force per unit area:

P=FAP=\frac{F}{A}

Common units include pascals, kilopascals, atmospheres, bars, and torr. A useful conversion is

1 atm=101.325 kPa=760 torr=1.01325 bar1\ \text{atm}=101.325\ \text{kPa}=760\ \text{torr}=1.01325\ \text{bar}

Temperature must be expressed on the absolute scale in gas-law calculations:

T(K)=T(∘C)+273.15T(\text{K})=T(^{\circ}\text{C})+273.15

For example, 25.0 ∘C=298.15 K25.0\ ^{\circ}\text{C}=298.15\ \text{K}, not 25.0 K25.0\ \text{K}.

Takeaway: Before selecting an equation, identify which variables change, which remain constant, and whether all units are compatible.

Choosing and Applying the Gas Laws

The empirical gas laws describe special cases in which all but two variables are held constant.

  • applies when amount and temperature are constant:

    P1V1=P2V2P_1V_1=P_2V_2

    Pressure increases when volume decreases.

  • applies when amount and pressure are constant:

    V1T1=V2T2\frac{V_1}{T_1}=\frac{V_2}{T_2}

    Volume increases with absolute temperature.

  • Amontons’s law applies when amount and volume are constant:

    P1T1=P2T2\frac{P_1}{T_1}=\frac{P_2}{T_2}

    Pressure increases with absolute temperature.

  • Avogadro’s law applies when pressure and temperature are constant:

    V1n1=V2n2\frac{V_1}{n_1}=\frac{V_2}{n_2}

    Volume increases with the amount of gas.

When the amount of gas is constant and pressure, volume, and temperature all change, use the combined gas law:

P1V1T1=P2V2T2\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

For example, a gas occupying 2.40 L2.40\ \text{L} at 1.20 atm1.20\ \text{atm} changes volume at constant temperature to a pressure of 2.00 atm2.00\ \text{atm}. gives

V2=(1.20 atm)(2.40 L)2.00 atm=1.44 LV_2=\frac{(1.20\ \text{atm})(2.40\ \text{L})}{2.00\ \text{atm}}=1.44\ \text{L}

Takeaway: Match the law to the variables held constant, and convert every temperature to kelvins before calculating.

The and Gas Quantities

The empirical relationships combine into the :

PV=nRTPV=nRT

Choose the value of RR that matches the pressure and volume units. Two useful forms are

R=0.082057 L⋅atmmol⋅KR=0.082057\ \frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}

and

R=8.31446 kPa⋅Lmol⋅KR=8.31446\ \frac{\text{kPa}\cdot\text{L}}{\text{mol}\cdot\text{K}}

To find the amount of gas, rearrange the equation:

n=PVRTn=\frac{PV}{RT}

For 5.00 L5.00\ \text{L} at 0.950 atm0.950\ \text{atm} and 27.0 ∘C27.0\ ^{\circ}\text{C}, first convert the temperature:

T=27.0+273.15=300.15 KT=27.0+273.15=300.15\ \text{K}

Then

n=(0.950 atm)(5.00 L)(0.082057 L⋅atm⋅mol−1⋅K−1)(300.15 K)=0.193 moln=\frac{(0.950\ \text{atm})(5.00\ \text{L})}{(0.082057\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1})(300.15\ \text{K})}=0.193\ \text{mol}

The equation also connects gas density dd and molar mass MM:

d=MPRTd=\frac{MP}{RT}

If density is known, molar mass can be found from

M=dRTPM=\frac{dRT}{P}

Takeaway: Write the desired variable first, check units, and use absolute temperature throughout.

Gas Mixtures and Pressure Contributions

In a gas mixture, each component contributes a . The mole fraction of component ii is

Xi=nintotalX_i=\frac{n_i}{n_{\text{total}}}

and its is

Pi=XiPtotalP_i=X_iP_{\text{total}}

The mole fractions sum to one:

∑Xi=1\sum X_i=1

For a mixture containing 2.0 mol2.0\ \text{mol} of nitrogen and 1.0 mol1.0\ \text{mol} of oxygen at a total pressure of 3.00 atm3.00\ \text{atm}, the total amount is 3.0 mol3.0\ \text{mol}. Therefore,

XN2=2.03.0=0.667,XO2=1.03.0=0.333X_{\mathrm{N_2}}=\frac{2.0}{3.0}=0.667,\qquad X_{\mathrm{O_2}}=\frac{1.0}{3.0}=0.333

so

PN2=(0.667)(3.00 atm)=2.00 atmP_{\mathrm{N_2}}=(0.667)(3.00\ \text{atm})=2.00\ \text{atm}

and

PO2=(0.333)(3.00 atm)=1.00 atmP_{\mathrm{O_2}}=(0.333)(3.00\ \text{atm})=1.00\ \text{atm}

When a gas is collected over water, the measured pressure includes water vapor. The dry-gas pressure is

Pdry gas=Ptotal−PH2OP_{\text{dry gas}}=P_{\text{total}}-P_{\mathrm{H_2O}}

For a total pressure of 755 torr755\ \text{torr} and water-vapor pressure of 23.8 torr23.8\ \text{torr}, the dry-gas pressure is 731 torr731\ \text{torr}.

Takeaway: Separate a mixture into components by mole fraction, and subtract water-vapor pressure when the goal is the pressure of a dry gas.

Particle Motion, Speed, and Transport

explains the gas laws through particle motion. An ideal gas contains many tiny particles that move continuously and randomly. Their own volume is negligible compared with the container volume, collisions are elastic, and intermolecular attractions are insignificant. The particles’ average kinetic energy increases with absolute temperature.

These ideas connect directly to the gas laws. Decreasing volume produces more frequent wall collisions, increasing pressure. Increasing temperature makes particles move faster; at constant pressure, the gas expands. Adding particles increases the number of wall collisions and, at constant pressure and temperature, increases volume.

Gas particles have a distribution of speeds rather than one common speed. Their root-mean-square speed is

urms=3RTMu_{\mathrm{rms}}=\sqrt{\frac{3RT}{M}}

where MM is molar mass in kilograms per mole. The speed increases with temperature and decreases with molar mass. The average translational kinetic energy per mole is

Ekin, mol=32RTE_{\text{kin, mol}}=\frac{3}{2}RT

Thus, ideal gases at the same temperature have the same average translational kinetic energy, even though lighter particles move faster on average.

Diffusion is the spreading of gas particles through space. Effusion is the escape of particles through a very small opening. describes the rate relationship:

r1r2=M2M1\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}

Takeaway: Temperature measures the scale of average particle kinetic energy, while molar mass affects how rapidly particles move.

Gas Stoichiometry and Nonideal Conditions

The ideal-gas model is most accurate at low pressure and high temperature. Under those conditions, particles are far apart, their individual volumes are relatively unimportant, and attractive forces have little effect. At high pressure, particle volume becomes more significant. At low temperature, attractive forces become more important, especially near condensation conditions.

For a gas reaction, connect particle-level amounts to measurable conditions through a consistent sequence:

  1. Write and balance the chemical equation.

  2. Convert gas measurements to moles with

    n=PVRTn=\frac{PV}{RT}
  3. Apply mole ratios from the balanced equation.

  4. Convert the resulting amount into the requested mass, volume, pressure, or temperature.

For example, calcium carbonate reacts with hydrochloric acid according to

CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g)\mathrm{CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2O(l)+CO_2(g)}

The mole ratio is 1 mol CaCO3:1 mol CO21\ \text{mol}\ \mathrm{CaCO_3}:1\ \text{mol}\ \mathrm{CO_2}. If 0.0250 mol0.0250\ \text{mol} of calcium carbonate reacts completely, then 0.0250 mol0.0250\ \text{mol} of carbon dioxide forms. At 298.15 K298.15\ \text{K} and 1.00 atm1.00\ \text{atm},

V=nRTP=(0.0250)(0.082057)(298.15)1.00=0.612 LV=\frac{nRT}{P}=\frac{(0.0250)(0.082057)(298.15)}{1.00}=0.612\ \text{L}

When gaseous substances react at the same temperature and pressure, their volume ratios equal the coefficients in the balanced equation. For

N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)+3H_2(g)\rightarrow 2NH_3(g)}

11 volume of nitrogen reacts with 33 volumes of hydrogen to form 22 volumes of ammonia. If reactant amounts do not match the required ratio, identify the . Compare gas volumes directly only when temperature and pressure are the same; otherwise, convert to moles first.

Takeaway: Balance first, convert to moles when necessary, apply stoichiometric ratios, and then return to the requested gas quantity.

Experimental Reliability and Error Checks

Reliable gas measurements require attention to both technique and calculation. Use calibrated instruments, check the apparatus for leaks, and allow the gas and its surroundings to reach thermal equilibrium. Record pressure, temperature, and volume with appropriate significant figures. Convert Celsius temperatures to kelvins and correct for water vapor when collecting gas over water.

To compare an experimental result with an accepted value, use percent error:

% error=∣experimental−acceptedaccepted∣×100%\%\ \text{error}=\left|\frac{\text{experimental}-\text{accepted}}{\text{accepted}}\right|\times100\%

A consistent difference between measured and predicted values may indicate a systematic error, such as a leak, an inaccurate pressure correction, or gas remaining dissolved in the collection liquid.

A concise calculation check is:

  • Are all temperatures in kelvins?

  • Do the pressure and volume units match the selected value of RR?

  • Has water-vapor pressure been subtracted when appropriate?

  • Does the result have sensible units and magnitude?

  • Were balanced-equation coefficients used as mole ratios?

Takeaway: Unit consistency, vapor-pressure corrections, equilibrium, and leak prevention are as important as the algebra.