DAT General Chemistry: Avogadro's Law and Molar Volume at STP

DAT High-Yield Concept

Avogadro's Law: At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles ( $V \propto n$ ). Equal volumes of any gas contain an equal number of particles.
STP Definition: Standard Temperature and Pressure for DAT calculations is defined as $\pu{0 degC}$ ( $\pu{273.15 K}$ ) and $\pu{1 atm}$ .
Molar Volume at STP: Exactly $\pu{1.00 mol}$ of any ideal gas occupies $\pu{22.4 L}$ at STP.
STP vs. Standard State: STP ( $\pu{0 degC}$ and $\pu{1 atm}$ ) is used for gas law problems. Standard State ( $\pu{25 degC}$ and $\pu{1 atm}$ ) is used for thermodynamic quantities ( $\Delta H^\circ$ , $\Delta S^\circ$ , $\Delta G^\circ$ ). Do not confuse their temperatures on the DAT!

Avogadro's Law

Avogadro's law states that at constant temperature ( $T$ ) and pressure ( $P$ ), the volume ( $V$ ) of a gas is directly proportional to the amount of gas in moles ( $n$ ). This relationship arises because gas molecules are separated by vast amounts of empty space under ideal conditions, meaning the physical size or identity of the gas molecule does not dictate the volume occupied (ideal gas behavior is covered in depth in its own lesson). Mathematically, Avogadro's law is written as:

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

Key Distinction: Avogadro's law applies to gas molecules regardless of molar mass. One mole of helium (

\ce{He}

\pu{4.00 g/mol}

) occupies the exact same volume as one mole of sulfur hexafluoride (

\ce{SF6}

\pu{146.06 g/mol}

) under the same temperature and pressure conditions.

Worked Example: Gas Addition at Constant Temperature and Pressure

Question: A flexible balloon contains $\pu{0.50 mol}$ of helium gas ( $\ce{He}$ ) in a volume of $\pu{11.2 L}$ . If $\pu{0.25 mol}$ of additional helium gas is injected into the balloon while maintaining constant temperature and pressure, what is the final volume of the balloon?

Solution:

Identify initial state variables: $n_1 = \pu{0.50 mol}$ , $V_1 = \pu{11.2 L}$ .
Calculate total final moles: $n_2 = n_1 + \Delta n = \pu{0.50 mol} + \pu{0.25 mol} = \pu{0.75 mol}$ .
Rearrange Avogadro's law to solve for $V_2$ : $V_2 = V_1 \times \left(\frac{n_2}{n_1}\right)$
Substitute values and solve using dimensional analysis: $V_2 = \pu{11.2 L} \times \frac{\pu{0.75 mol}}{\pu{0.50 mol}} = \pu{16.8 L}$

Exam Trap: DAT questions often state that a specific amount of gas is added. Students mistakenly set $n_2$ equal to the added amount ( $\pu{0.25 mol}$ ) rather than the total final amount ( $n_1 + \Delta n = \pu{0.75 mol}$ ). Always sum the initial and added moles to find $n_2$ .

Standard Temperature and Pressure (STP) and Molar Volume

Standard Temperature and Pressure (STP) provides a universal baseline for reporting gas measurements. On the DAT, STP is defined as a temperature of $\pu{0 degC}$ ( $\pu{273.15 K}$ ) and a pressure of $\pu{1.00 atm}$ . Substituting these standard conditions into the ideal gas equation yields the standard molar volume of an ideal gas:

V_m = \frac{RT}{P} = \frac{(\pu{0.08206 L*atm/(mol*K)})(\pu{273.15 K})}{\pu{1.00 atm}} = \pu{22.4 L/mol}

Mnemonic: "At STP, 22.4 fills the floor!" One mole of any ideal gas occupies

\pu{22.4 L}

\pu{0 degC}

and

\pu{1 atm}

One mole of any gas fills the same volume at 0 C and 1 atm — At the same temperature and pressure, equal volumes of gases hold equal numbers of molecules, so one mole of any gas that behaves ideally occupies about 22.4 L at 0 degrees Celsius and 1 atm. The three boxes therefore hold the same number of particles even though helium, nitrogen and carbon dioxide have very different molar masses. Real gases follow this closely near these conditions but not exactly.

Worked Example: Molar Volume Calculation at STP

Question: What volume does $\pu{44.0 g}$ of carbon dioxide gas ( $\ce{CO2}$ ) occupy at STP?

Solution:

Find the molar mass of $\ce{CO2}$ : $12.01 + 2(16.00) = \pu{44.01 g/mol}$ .
Convert mass of $\ce{CO2}$ to moles: $n = \pu{44.0 g} \times \frac{\pu{1 mol}}{\pu{44.01 g}} = \pu{1.00 mol}$
Use the molar volume conversion factor at STP: $V = \pu{1.00 mol} \times \frac{\pu{22.4 L}}{\pu{1 mol}} = \pu{22.4 L}$

Exam Trap: Never apply the $\pu{22.4 L/mol}$ conversion factor to liquids, solids, or solutions! It is strictly valid for gases at STP. Furthermore, if conditions depart from STP (e.g., $\pu{25 degC}$ ), you cannot use $\pu{22.4 L/mol}$ .

Standard Temperature and Pressure (STP) vs. Standard State

Pre-dental students often confuse STP with Standard State. While STP establishes reference conditions for gases, Standard State sets reference conditions for thermochemistry, chemical equilibria, and electrochemistry.

Property	STP (Gas Laws)	Standard State (Thermodynamics)
Temperature	$\pu{0 degC}$ ( $\pu{273.15 K}$ )	$\pu{25 degC}$ ( $\pu{298.15 K}$ )
Pressure	$\pu{1 atm}$	$\pu{1 atm}$
Solution Concentration	Not applicable	$\pu{1.0 M}$
Gas Molar Volume	$\pu{22.4 L/mol}$	$\pu{24.5 L/mol}$
DAT Context	Gas stoichiometry, Avogadro's law	$\Delta H^\circ$ , $\Delta S^\circ$ , $\Delta G^\circ$ , $\mathcal{E}^\circ$

Critical Thinking Questions

A rigid container holds $\pu{1.0 mol}$ of nitrogen gas ( $\ce{N2}$ ) and another identical rigid container holds $\pu{1.0 mol}$ of chlorine gas ( $\ce{Cl2}$ ) at STP. Compare their volumes, total particle counts, and masses.
Explain why the molar volume of an ideal gas at Standard State ( $\pu{25 degC}$ , $\pu{1 atm}$ ) is larger than its molar volume at STP ( $\pu{0 degC}$ , $\pu{1 atm}$ ). Which gas law predicts this change?
If a reaction yields $\pu{0.50 mol}$ of oxygen gas ( $\ce{O2}$ ) collected over water at $\pu{0 degC}$ and a total pressure of $\pu{1.00 atm}$ , why is the dry volume of $\ce{O2}$ slightly less than $\pu{11.2 L}$ ?

References

Sources returned by Google Search grounding during lesson verification.

8.9: Avogadro's Law - The Relation between Volume and Molar Amount - Chemistry LibreTexts — chem.libretexts.org
Standard temperature and pressure - Wikipedia — en.wikipedia.org

Avogadro's Law and Molar Volume at STP

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