# Nakafa Framework: LLM

> For AI agents: use [llms.txt](https://nakafa.com/llms.txt). Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`.

URL: https://nakafa.com/en/subject/high-school/10/chemistry/basic-chemistry-laws/combining-volumes-law
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/10/chemistry/basic-chemistry-laws/combining-volumes-law/en.mdx

Output docs content for large language models.

---

import { CombiningVolumesLab } from "@repo/design-system/components/contents/chemistry/combining-volumes-law/lab";

export const metadata = {
  title: "Combining Volumes",
  description:
    "Learn how to read gas volume ratios from reaction coefficients when temperature and pressure are kept the same.",
  authors: [{ name: "Nabil Akbarazzima Fatih" }],
  date: "05/02/2026",
  subject: "Basic Laws of Chemistry",
};

## Gas Volumes Follow Coefficients

The law of combining volumes states that reacting gases, and gaseous products, have volumes in simple whole-number ratios when measured at the same temperature and pressure. This law is also known as Gay-Lussac's law of combining gas volumes.

OpenStax Chemistry 2e explains that gas volume ratios in a reaction are given by the coefficients in the balanced equation when all gas volumes are measured at the same temperature and pressure in [Stoichiometry of Gaseous Substances, Mixtures, and Reactions](https://openstax.org/books/chemistry-2e/pages/9-3-stoichiometry-of-gaseous-substances-mixtures-and-reactions). Britannica also summarizes Gay-Lussac's finding that hydrogen and oxygen combine by volume in a <InlineMath math="2:1" /> ratio to form water in [Gay-Lussac's law of combining volumes](https://www.britannica.com/science/Gay-Lussacs-law-of-combining-volumes).

The important word is **gas**. Coefficients can be read directly as volume ratios only for substances in the gas phase and only when those volumes are compared at the same temperature and pressure.

## Read Water as Vapor

Look at the balanced equation for forming water vapor.

<BlockMath math="2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(g)}" />

The coefficients <InlineMath math="2:1:2" /> mean <InlineMath math="2" /> volume parts of hydrogen react with <InlineMath math="1" /> volume part of oxygen to form <InlineMath math="2" /> volume parts of water vapor.

<BlockMath math="V_{\mathrm{H_2}} : V_{\mathrm{O_2}} : V_{\mathrm{H_2O}} = 2:1:2" />

So, if every gas is measured at the same temperature and pressure, <InlineMath math="200\ \mathrm{mL}" /> of hydrogen reacts with <InlineMath math="100\ \mathrm{mL}" /> of oxygen to produce <InlineMath math="200\ \mathrm{mL}" /> of water vapor. If the water has condensed into liquid, the liquid volume is no longer read directly from this gas ratio.

## Gas Volume Meter

Choose a reaction below. Each tube represents one volume part of gas at the same temperature and pressure.

<CombiningVolumesLab
  title={<>Gas Volume Meter</>}
  description={
    <>
      Compare tube heights to see reaction coefficients as gas volume ratios.
    </>
  }
  labels={{
    chooseMode: "Choose gas reaction",
    reactants: "Reactants",
    products: "Products",
    reactionView: "Law of combining volumes gas model",
    ratioLabel: "Volume ratio",
    exampleLabel: "Scaled example",
    volumeUnit: "vol",
    modes: {
      "water-vapor": {
        tab: "Steam",
        tabLabel: "Water vapor formation",
        helperCaption: (
          <>
            Two parts of <InlineMath math="\mathrm{H_2}" /> meet one part of{" "}
            <InlineMath math="\mathrm{O_2}" /> and form two parts of{" "}
            <InlineMath math="\mathrm{H_2O(g)}" />.
          </>
        ),
        ratio: (
          <>
            <InlineMath math="\mathrm{H_2:O_2:H_2O} = 2:1:2" />.
          </>
        ),
        example: (
          <>
            <InlineMath math="200\ \mathrm{mL}" /> of{" "}
            <InlineMath math="\mathrm{H_2}" /> needs{" "}
            <InlineMath math="100\ \mathrm{mL}" /> of{" "}
            <InlineMath math="\mathrm{O_2}" />.
          </>
        ),
      },
      "ammonia-synthesis": {
        tab: "Ammonia",
        tabLabel: "Ammonia formation",
        helperCaption: (
          <>
            One part of <InlineMath math="\mathrm{N_2}" /> needs three parts of{" "}
            <InlineMath math="\mathrm{H_2}" /> to form two parts of{" "}
            <InlineMath math="\mathrm{NH_3}" />.
          </>
        ),
        ratio: (
          <>
            <InlineMath math="\mathrm{N_2:H_2:NH_3} = 1:3:2" />.
          </>
        ),
        example: (
          <>
            <InlineMath math="1\ \mathrm{L}" /> of{" "}
            <InlineMath math="\mathrm{N_2}" /> needs{" "}
            <InlineMath math="3\ \mathrm{L}" /> of{" "}
            <InlineMath math="\mathrm{H_2}" />.
          </>
        ),
      },
      "ammonia-decomposition": {
        tab: "Decompose",
        tabLabel: "Ammonia decomposition",
        helperCaption: (
          <>
            Two parts of <InlineMath math="\mathrm{NH_3}" /> break into three
            parts of <InlineMath math="\mathrm{H_2}" /> and one part of{" "}
            <InlineMath math="\mathrm{N_2}" />.
          </>
        ),
        ratio: (
          <>
            <InlineMath math="\mathrm{NH_3:H_2:N_2} = 2:3:1" />.
          </>
        ),
        example: (
          <>
            <InlineMath math="600\ \mathrm{mL}" /> of{" "}
            <InlineMath math="\mathrm{NH_3}" /> produces{" "}
            <InlineMath math="900\ \mathrm{mL}" /> of{" "}
            <InlineMath math="\mathrm{H_2}" /> and{" "}
            <InlineMath math="300\ \mathrm{mL}" /> of{" "}
            <InlineMath math="\mathrm{N_2}" />.
          </>
        ),
      },
    },
  }}
/>

The model is not comparing mass. It compares gas volume. Because the temperature and pressure are the same, each volume part follows a coefficient in the balanced reaction.

## Ammonia Decomposition Example

Suppose <InlineMath math="600\ \mathrm{mL}" /> of ammonia gas decomposes completely at the same temperature and pressure. The balanced equation is:

<BlockMath math="2\mathrm{NH_3(g)} \rightarrow 3\mathrm{H_2(g)} + \mathrm{N_2(g)}" />

The coefficients give this volume ratio:

<BlockMath math="V_{\mathrm{NH_3}} : V_{\mathrm{H_2}} : V_{\mathrm{N_2}} = 2:3:1" />

Because <InlineMath math="2" /> parts of ammonia equal <InlineMath math="600\ \mathrm{mL}" />, <InlineMath math="1" /> volume part equals <InlineMath math="300\ \mathrm{mL}" />.

<BlockMath math="\begin{aligned}
V_{\mathrm{H_2}} &= 3 \times 300 = 900\ \mathrm{mL} \\
V_{\mathrm{N_2}} &= 1 \times 300 = 300\ \mathrm{mL}
\end{aligned}" />

So decomposing <InlineMath math="600\ \mathrm{mL}" /> of ammonia gas produces <InlineMath math="900\ \mathrm{mL}" /> of hydrogen gas and <InlineMath math="300\ \mathrm{mL}" /> of nitrogen gas.

## Where the Shortcut Stops

This law is useful, but its conditions must stay visible.

- The reaction equation must already be balanced.
- The substances being compared must be gases.
- All gas volumes must be measured at the same temperature and pressure.
- If a product turns into a liquid or solid, that product volume is not read directly from the gas coefficients.

With those conditions in place, reaction coefficients do more than count moles. For gases under the same conditions, they also give a quick map of relative volumes.