Q.Prove Mayer's equation: Cp - Cv = R. OR What is an adiabatic process? Determine the formula for work done in an adiabatic process.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Heat Capacity at Constant Pressure
Heat Capacity at Constant Pressure — From Intuition to Precision
Imagine you have a pot of water on a stove. You turn the burner on, and the water gets hotter. How much heat does it take to raise its temperature by, say, 10°C? That depends on two things: how much water you have, and whether the pot is open to the air or sealed tight.
If the pot is open (constant pressure — the air above it is always at atmospheric pressure), the water can expand as it heats. Some of the energy you supply goes into pushing the atmosphere aside — doing work against the outside air. So you need to put in more heat than if the pot were sealed (constant volume), where no expansion work is possible.
That extra heat is the key idea behind heat capacity at constant pressure, denoted Cp.
The Intuition First
Heat capacity tells you: "How much heat must I add to raise the temperature of this substance by 1°C (or 1 K)?"
- At constant volume (Cv): All the heat goes into increasing the internal energy (the kinetic and potential energy of the molecules). No work is done because the volume doesn't change.
- At constant pressure (Cp): Some heat goes into internal energy, but some also goes into the work of expansion against the constant external pressure. So Cp is always larger than Cv for gases (and for most solids/liquids, the difference is tiny because they barely expand).
For an ideal gas, the difference is exactly Cp−Cv=nR, where n is the number of moles and R is the universal gas constant. This is a direct consequence of the first law of thermodynamics.
The Precise Statement
Heat capacity at constant pressure is defined as the amount of heat required to raise the temperature of a substance by 1 K (or 1°C) while keeping the pressure constant.
Mathematically:
Cp=(dTδQ)p
The subscript p means "at constant pressure." The δQ (not dQ) reminds us that heat is a path-dependent quantity, not a state function.
But we can rewrite this in terms of a state function — enthalpy (H). At constant pressure, the heat added equals the change in enthalpy:
δQp=dH
Therefore:
Cp=(∂T∂H)p
This is the working definition you'll use in problems: Cp is the partial derivative of enthalpy with respect to temperature at constant pressure.
Molar vs. Specific Heat Capacity
You'll encounter two common forms:
- Molar heat capacity at constant pressure (Cp,m): heat capacity per mole (units: J mol⁻¹ K⁻¹)
- Specific heat capacity at constant pressure (cp): heat capacity per unit mass (units: J kg⁻¹ K⁻¹)
The total heat capacity of a sample is:
Cp=n⋅Cp,m=m⋅cp
Why It Matters
In most chemical reactions and physical processes, the system is open to the atmosphere — constant pressure. So Cp is the relevant quantity for:
- Calculating enthalpy changes (ΔH=nCp,mΔT)
- Designing calorimeters (like coffee-cup calorimeters that operate at constant pressure) …
Applying the first law of thermodynamics separately to one mole of an ideal gas heated at constant volume and at constant pressure shows that the extra heat needed per degree rise at constant pressure exactly equals the work done in expansion, R dT, which leads directly to Mayer's relation Cp minus Cv equals R. …
For one mole of an ideal gas, Cp − Cv = R, because heat supplied at constant pressure must also do expansion work, unlike heat supplied at constant volume.
Consider one mole of an ideal gas, and apply the first law of thermodynamics, dQ = dU + dW, to two separate processes that each raise the temperature by the same small amount dT:
Case 1 — Heating at constant volume: Since volume is constant, no work is done, dW = 0. So:
dQ_v = dU
By definition, Cv = dQ_v/dT, so:
Cv dT = dU ... (i)
Case 2 — Heating at constant pressure: Here the gas expands as it is heated, doing work dW = P dV. So:
dQ_p = dU + P dV
By definition, Cp = dQ_p/dT, so:
Cp dT = dU + P dV ... (ii)
For an ideal gas, PV = RT (one mole). At constant pressure, differentiating: P dV = R dT.
Substituting into (ii):
Cp dT = dU + R dT
But from (i), dU = Cv dT. Substituting:
Cp dT = Cv dT + R dT
Dividing through by dT: …
Showing the 12 most recent of 14 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The ratio of Molar specific Heats (Cp/Cv) of Monoatomic gas is :(a) 5/3(b) 5/7(c) 7/9(d) 7/5.
›Reveal solutionSolution
A monoatomic gas has 3 degrees of freedom, giving Cv = (3/2)R, Cp = (5/2)R, and hence Cp/Cv = 5/3.
By the law of equipartition of energy, each degree of freedom contributes (1/2)RT to the internal energy per mole of an ideal gas.
A monoatomic gas (e.g. He, Ar) has only 3 translational degrees of freedom (no rotational or vibrational contribution for a point-like molecule at ordinary temperatures).
…
- CBSE 2026Set ANNUAL1 markMCQQ.The ratio of Cp and Cv for helium gas is(a) 5/3(b) 5/7(c) 3/5(d) 9/7
›Reveal solutionSolution
Helium is monatomic, so gamma = Cp/Cv = 5/3. Answer (A).
For a monatomic ideal gas (3 translational degrees of freedom):
Cv = (3/2)R and Cp = Cv + R = (5/2)R.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Mayer's formula is (A) C_P - C_V = R (B) C_P/C_V = R (C) C_V - C_P = R (D) C_V/C_P = R
›Reveal solutionSolution
Mayer's formula states CP−CV=R for one mole of an ideal gas.
At constant volume, all the heat supplied goes into raising internal energy: dQ=CVdT=dU.
At constant pressure, the gas also does work as it expands, so extra heat is needed for the same temperature rise: dQ=CPdT=dU+PdV.
For 1 mole of an ideal gas, PV=RT⇒PdV=RdT at constant pressure. Substituting:
…
- CBSE 2025Set ANNUAL1 markQ.Answer in one word or one sentence: Write Mayer's equation.
›Reveal solutionSolution
Mayer's relation states Cp - Cv = R for an ideal gas.
For one mole of an ideal gas, the molar specific heat at constant pressure (Cp) exceeds the molar specific heat at constant volume (Cv), because at constant pressure some of the supplied heat must also do work as the gas expands, in addition to raising its internal energy. Using the first law of thermodynamics and the ideal gas e …
- CBSE 2024Set ANNUAL1 markMCQQ.The ratio of specific heats (γ) of diatomic gas is:(a) 3/5(b) 5/7(c) 7/9(d) 7/5
›Reveal solutionSolution
Using the equipartition theorem, a diatomic gas has 5 degrees of freedom, giving Cv = 5R/2, Cp = 7R/2, and hence gamma = Cp/Cv = 7/5.
For an ideal gas molecule with f degrees of freedom, the equipartition theorem gives the molar internal energy as U = (f/2)RT, so:
Cv = dU/dT = (f/2)R
Cp = Cv + R = (f/2)R + R = [(f+2)/2]R
gamma = Cp/Cv = (f+2)/f
…
- CBSE 2023Set ANNUAL1 markMCQQ.Mayer's formula for the relation between two principal specific heats Cp and Cv of a gas is given by(1) Cv - Cp = R(2) Cp / Cv = R(3) Cp - Cv = R(4) Cv / Cp = R
›Reveal solutionSolution
For one mole of an ideal gas, Mayer's relation states Cp - Cv = R, where R is the universal gas constant.
For an ideal gas, heating it at constant volume only increases its internal energy (all the heat goes into raising temperature), while heating it at constant pressure must also supply energy for the gas to do work as it expands (W = PΔV = RΔT for one mole). This extra work requirement is exactly what makes Cp larger than Cv, and the difference between them equals the gas constant R per mole:
Cp - Cv = R
…
- CBSE 2022Set ANNUAL1 markMCQQ.In the relation PV^γ = constant, γ is:(a) Cp - Cv(b) Cp / Cv(c) Cp . Cv(d) Cp + Cv
›Reveal solutionSolution
The symbol γ in the adiabatic relation PV^γ = constant is the ratio of specific heat at constant pressure to specific heat at constant volume, γ = Cp/Cv.
For a reversible adiabatic process on an ideal gas, combining the first law of thermodynamics (dQ = 0) with the ideal gas equation PV = nRT gives PV^γ = constant, where γ = Cp/Cv. This ratio is always greater than 1 because Cp > Cv (at constant pressure, some of the added heat goes int …
- CBSE 2022Set ANNUAL1 markMCQQ.Match Column A item "Mayer's relation" with the correct item in Column B.(a) kg m^2(b) Poise(c) Cp - Cv = R(d) E = mc^2(e) 1/frequency(f) distance(g) 24 hours
›Reveal solutionSolution
Mayer's relation states Cp - Cv = R (per mole), showing that the extra heat needed at constant pressure over constant volume goes entirely into the work of expansion.
…
- CBSE 2022Set ANNUAL1 markMCQQ.(Cp − Cv) is equal to —(a) R x J(b) R − J(c) R/J(d) J/R
›Reveal solutionSolution
Cp − Cv = R/J (molar heat capacities in calorie/heat units).
Mayer's relation for one mole of an ideal gas, when the heat capacities are in energy (joule) units, is Cp−Cv=R.
…
- CBSE 2020Set ANNUAL1 markMCQQ.The molar specific heat at constant pressure of an ideal gas is (7/2) R. The ratio of specific heat at constant pressure to that at constant volume is:(a) 9/7(b) 7/5(c) 8/7(d) 5/7
›Reveal solutionSolution
Mayer's relation connects Cp and Cv for an ideal gas (Cp - Cv = R); once Cv is found, the ratio gamma = Cp/Cv follows directly.
Given Cp = (7/2) R.
Mayer's relation: Cp - Cv = R, so Cv = Cp - R = (7/2) R - R = (5/2) R.
gamma = Cp / Cv = [(7/2) R] / [(5/2) R] = 7/5.
…
- CBSE 2020Set ANNUAL1 markMCQQ.The kinetic energy of one mole of an ideal gas is E = 3/2 RT. The value of Cp is:(a) 0.5 R(b) 0.1 R(c) 1.5 R(d) 2.5 R
›Reveal solutionSolution
The given internal (kinetic) energy expression identifies Cv = (3/2)R directly; adding R (Mayer's relation) gives Cp = (5/2)R = 2.5R.
For one mole of an ideal gas, internal energy (here treated as the total kinetic energy) is E = (3/2) R T, which by definition means:
Cv = dE/dT = (3/2) R.
…
- CBSE 2020Set ANNUAL1 markMCQQ.If S_P and S_V denote the specific heats of nitrogen gas per unit mass at constant pressure and constant volume respectively, then:(a) S_P - S_V = 28R(b) S_P - S_V = R/28(c) S_P - S_V = R/14(d) S_P - S_V = R
›Reveal solutionSolution
Mayer's relation C_P - C_V = R applies to molar specific heats; converting to specific heat per unit mass by dividing by the molar mass of nitrogen (28 g/mol) gives S_P - S_V = R/28.
Mayer's relation for an ideal gas relates the molar specific heats at constant pressure and constant volume:
C_P - C_V = R
…
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