B.3.4—Empirical gas laws

One relation for a fixed amount of gas

The constant-pressure, constant-volume and constant-temperature observations combine to give

$\frac{PV}{T}=\text{constant}$

provided the amount of gas is unchanged.

Compare two states

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Temperatures must be absolute. If pressure and temperature stay fixed, volume is fixed; if volume and temperature stay fixed, pressure is fixed.

Use the constraint first

Name what is constant before choosing a simplified law: Boyle-type behaviour uses constant T, Charles-type behaviour uses constant P, and pressure–temperature behaviour uses constant V.

Worked example from the mapped local textbook

A fixed gas changes from $P_1=1.1\times10^5\,\mathrm{Pa}$ , $V_1=0.27\,\mathrm{m^3}$ and $T_1=289\,\mathrm{K}$ to $V_2=0.35\,\mathrm{m^3}$ and $T_2=423\,\mathrm{K}$ .

$P_2=\frac{P_1V_1T_2}{T_1V_2}=\frac{(1.1\times10^5)(0.27)(423)}{(289)(0.35)}=1.2\times10^5\,\mathrm{Pa}$

The amount of gas must remain fixed for this two-state relation.

Common trap

Do not apply $PV/T=\text{constant}$ after gas has been added or removed unless the problem explicitly tracks the changed amount of gas.

How it is assessed

The evidence tests pressure after joining containers at constant temperature and a temperature calculation for an isobaric process.

Command terms

Determine / Show

What earns marks

For fixed amount of gas use P₁V₁/T₁=P₂V₂/T₂. Identify the constant process before simplifying; use kelvin and preserve units. For connected containers at the same temperature, conserve total PV and divide by total volume.

Watch for

Changing the gas amount without accounting for it, or using Celsius in the combined gas law.

Representative question

[Maximum number: 3]

Two containers of volume $0.20 \mathrm{~m}^{3}$ and $0.10 \mathrm{~m}^{3}$ are filled with an ideal gas. The pressure in the larger container is $3.0 \times 10^{4} \mathrm{~Pa}$ . The pressure in the smaller container is $9.0 \times 10^{4} \mathrm{~Pa}$ . The temperature of the gas in both containers is the same. A thin tube with a valve joins the containers. The valve is initially closed.

The valve is opened so that gas can move from one container to the other. The temperature remains unchanged.

Determine the new pressure of the gas.

ALTERNATIVE 1

pressure due to gas in left container $3.0 \times 10^{4} \times \frac{0.20}{0.30}=2.0 \times 10^{4}$ «Pa» pressure due to gas in right container $9.0 \times 10^{4} \times \frac{0.10}{0.30}=3.0 \times 10^{4}$ «Ра» adding gives $P=5.0 \times 10^{4}$ «Ра»

ALTERNATIVE 2

number of moles <<in a container is>> $\frac{3.0 \times 10^{4} \times 0.20}{R T} \boldsymbol{OR} \frac{9.0 \times 10^{4} \times 0.10}{R T}$ $P \times 0.30=\left(\frac{3.0 \times 10^{4} \times 0.20}{R T}+\frac{9.0 \times 10^{4} \times 0.10}{R T}\right) R T$ $P=5.0 \times 10^{4} \ll \mathrm{~Pa} \gg$

ALTERNATIVE 3

Use of $P_{1} V_{1}+P_{2} V_{2}=P\left(V_{1}+V_{2}\right)$ $3.0 \times 10^{4} \times 0.2+9.0 \times 10^{4} \times 0.1=P(0.2+0.1)$ $P=5.0 \times 10^{4} \ll P a \gg$

Macroscopic equations

Pressure is $P=F_{\perp}/A$ . For a fixed amount of gas, empirical laws combine to $PV/T=\text{constant}$ , and the ideal-gas equations are $PV=nRT=Nk_BT$ .

Microscopic model

Particles move randomly and collide elastically with walls. Momentum transfer produces pressure, with $P=\frac13\rho\overline{v^2}$ . For a monatomic ideal gas, $U=\frac32Nk_BT=\frac32nRT$ .

Bridge the descriptions

Use $n=N/N_A$ to move between moles and particles. Choose the equation from the data provided, convert temperature to kelvin, and keep SI units consistent.

Model boundary

The ideal approximation works best at high temperature and low pressure or density. At high density, high pressure or near condensation, finite particle size and intermolecular forces matter.

B.3.4—Empirical gas laws

Combine the Empirical Gas Laws

B.3.4 Exam Analysis

Assessment in practice

Synthesize B.3 Gas Laws

Question 1