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Mayer's relation

Mayer's relation is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Mayer's relation rather than just read about it. In short: In the 19th century, German chemist and physicist Julius von Mayer derived a relation between the molar heat capacity at constant pressure and the molar heat capacity at constant volume for an ideal gas. Mayer's relation states that C P , m − C V , m = R , {\displaystyle C_{P,\mathrm {m} }-C_{V,\mathrm {m} }=R,} where CP,m is the molar heat capacity at constant pressure, CV,m is the molar heat capacity at constant v…

Key takeaways

  • Mayer's relation belongs to mathematics; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Mayer's relation to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Mayer's relation from memory before moving on to harder problems.

Reference excerpt

In the 19th century, German chemist and physicist Julius von Mayer derived a relation between the molar heat capacity at constant pressure and the molar heat capacity at constant volume for an ideal gas. Mayer's relation states that

C P , m − C V , m = R , {\displaystyle C_{P,\mathrm {m} }-C_{V,\mathrm {m} }=R,}

where CP,m is the molar heat capacity at constant pressure, CV,m is the molar heat capacity at constant volume and R is the gas constant.

General form For more general homogeneous substances, not just ideal gases, the difference takes the form,

C P , m − C V , m = V m T α V 2 β T {\displaystyle C_{P,\mathrm {m} }-C_{V,\mathrm {m} }=V_{\mathrm {m} }T{\frac {\alpha _{V}^{2}}{\beta _{T}}}}

(see relations between heat capacities), where V m {\displaystyle V_{\mathrm {m} }} is the molar volume, T {\displaystyle T} is the temperature, α V {\displaystyle \alpha _{V}} is the thermal expansion coefficient and β {\displaystyle \beta } is the isothermal compressibility. From this latter relation, several inferences can be made:

Since the isothermal compressibility β T {\displaystyle \beta _{T}} is positive for nearly all phases, and the square of thermal expansion coefficient α {\displaystyle \alpha } is always either a positive quantity or zero, the specific heat at constant pressure is nearly always greater than or equal to specific heat at constant volume: C P , m ≥ C V , m . {\displaystyle C_{P,\mathrm {m} }\geq C_{V,\mathrm {m} }.} There are no known exceptions to this principle for gases or liquids, but certain solids are known to exhibit negative compressibilities and presumably these would be (unusual) cases where C P , m < C V , m {\displaystyle C_{P,\mathrm {m} }<C_{V,\mathrm {m} }} . For incompressible substances, CP,m and CV,m are identical. Also for substances that are nearly incompressible, such as solids and liquids, the difference between the two specific heats is negligible. As the absolute temperature of the system approaches zero, since both heat capacities must generally approach zero in accordance with the Third Law of Thermodynamics, the difference between CP,m and CV,m also approaches zero. Exceptions to this rule might be found in systems exhibiting residual entropy due to disorder within the crystal.

References

Worked examples

Example 1 — a first encounter with Mayer's relation

Start with the simplest possible case. Write down what Mayer's relation claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Mayer's relation before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Mayer's relation ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Mayer's relation

In research
Mayer's relation appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Mayer's relation in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Mayer's relation is common in secondary-school and first-year university syllabi. It links to neighbouring topics Thermodynamic equations, so understanding it makes those chapters shorter.
In everyday life
Look for Mayer's relation outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

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How to study Mayer's relation in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Mayer's relation means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Mayer's relation out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Mayer's relation in simple terms?

In the 19th century, German chemist and physicist Julius von Mayer derived a relation between the molar heat capacity at constant pressure and the molar heat capacity at constant volume for an ideal gas. Mayer's relation states that C P , m − C V , m = R , {\displaystyle C_{P,\mathrm {m} }-C_{V,\ma…

Why does Mayer's relation matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Mayer's relation?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Mayer's relation.

Tags

  • Thermodynamic equations

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