Specific heat capacity

Put a sealed cylinder of gas over a flame and its temperature climbs steadily — but ask "how much heat did that take?" and the honest answer is: it depends. Not on the gas alone, and not on the temperature rise alone, but on whether the piston above the gas was locked in place or free to rise. Heat a solid block and this question barely matters, because solids and liquids hardly change volume when warmed. Heat a gas, and the answer can differ by nearly forty per cent depending on that one condition — because a gas that is free to expand spends part of the incoming heat pushing its surroundings out of the way, and only the rest goes into actually raising its temperature. This is why, unlike a solid or a liquid, a gas does not have one specific heat capacity — it has two, $C_V$ at constant volume and $C_P$ at constant pressure, and the gap between them is not some fudge factor but a clean consequence of the first law of thermodynamics. Chase that gap with nothing more than $\Delta Q = \Delta U + \Delta W$ and the ideal gas law $PV = RT$ , and it collapses to one of the tidiest results in the whole syllabus: $C_P - C_V = R$ , Mayer's relation, exact for every ideal gas regardless of what it is made of. This article builds $C_V$ and $C_P$ from the first law, derives Mayer's relation step by step, and works two numeric examples — one at board level, one competitive — before laying out the traps that lose the most marks: assuming $C_P = C_V$ out of solid-and-liquid habit, muddling molar specific heat with specific heat per unit mass, and forgetting that $\Delta U$ for an ideal gas depends on temperature alone.

Key takeaways

A gas needs two specific heats because heating it can happen at constant volume or constant pressure, and these require different amounts of heat for the same temperature rise.
$C_V$ is the molar heat capacity at constant volume: since $\Delta V = 0$ , no work is done, so all the heat raises internal energy, $C_V = (\Delta U/\Delta T)_V$ .
$C_P$ is the molar heat capacity at constant pressure: the gas also expands and does work $P\Delta V$ , so extra heat is needed for the same $\Delta T$ .
Because an ideal gas's internal energy depends only on temperature, $\Delta U$ for a given $\Delta T$ is identical whether $V$ or $P$ is held fixed — this is the key step that unlocks the derivation.
Applying the first law at constant pressure with $P\Delta V = R\Delta T$ (for one mole) gives Mayer's relation, $C_P - C_V = R$ .
The ratio $\gamma = C_P/C_V$ is always greater than 1 and appears again in the adiabatic relation $PV^\gamma = \text{constant}$ , later in this chapter; its numeric value comes from degrees of freedom, covered in the Kinetic Theory unit.
Key idea: for an ideal gas, $C_P$ always exceeds $C_V$ by exactly $R$ per mole, because constant-pressure heating must additionally supply the work $P\Delta V = R\Delta T$ that constant-volume heating does not.

The core insight

For one mole of an ideal gas, $C_P - C_V = R$ because constant-pressure heating must supply, on top of the same rise in internal energy as constant-volume heating, the extra work $P\Delta V = R\Delta T$ done by the expanding gas.

Specific heat capacity (of gases)

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics Thermodynamics unit, immediately after the first law of thermodynamics ( $\Delta Q = \Delta U + \Delta W$ , with $\Delta W = P\Delta V$ ) and before the chapter moves on to thermodynamic state variables and the classification of processes (isothermal, adiabatic, cyclic). You are expected to define $C_V$ and $C_P$ for an ideal gas, derive Mayer's relation $C_P - C_V = R$ using the first law, and know that $\gamma = C_P/C_V$ is the ratio used later in the adiabatic relation $PV^\gamma = \text{const}$ . The numeric values of $C_V$ and $C_P$ that come from a gas's degrees of freedom — $\tfrac32R$ for monatomic, $\tfrac52R$ for diatomic — are developed properly in the Kinetic Theory unit's law of equipartition of energy; here they are only quoted and used, never derived from molecular structure. Nothing in this article strays into the out-of-scope territory of heat engines or entropy.

The physical picture

Start from the first law of thermodynamics for a small change: heat supplied equals the rise in internal energy plus the work done by the gas, $\Delta Q = \Delta U + \Delta W$ , with $\Delta W = P\Delta V$ for a gas pushing against a piston at pressure $P$ . This single equation is the whole engine behind everything below — the only new ingredient is asking what happens when you hold one variable, $V$ or $P$ , fixed while heating.

Molar specific heat at constant volume. Lock the piston so the gas cannot expand. Then $\Delta V = 0$ , so $\Delta W = 0$ , and every joule of heat goes straight into internal energy:

\Delta Q = \Delta U \quad\Rightarrow\quad C_V = \left(\frac{\Delta U}{\Delta T}\right)_V

where: $C_V$ is the molar heat capacity at constant volume (J mol $^{-1}$ K $^{-1}$ ), $\Delta U$ is the rise in internal energy, and $\Delta T$ is the temperature rise. Physically, $C_V$ measures how "expensive" in heat it is to raise the gas's internal energy alone, with no expansion work to pay for.

Molar specific heat at constant pressure. Now free the piston and let the gas push it out as it warms, at fixed pressure $P$ . The gas still needs the same rise in internal energy for the same $\Delta T$ — but it must also supply the work of pushing the piston, so more heat is needed overall:

C_P = \left(\frac{\Delta Q}{\Delta T}\right)_P

where: $C_P$ is the molar heat capacity at constant pressure (J mol $^{-1}$ K $^{-1}$ ). The physical reading is that $C_P$ bundles two costs together — raising the internal energy, exactly as before, plus paying for expansion work — while $C_V$ pays only the first cost.

Figure

Two identical cylinders of gas being heated by the same amount, the left one with a fixed piston showing all the heat entering as a rise in internal energy only, the right one with a free piston that rises as the gas expands, showing the same rise in internal energy plus extra heat spent as work pushing the piston up

The crucial fact linking the two cases is that internal energy of an ideal gas depends only on temperature — not on volume or pressure separately. So for the same $\Delta T$ , $\Delta U$ is identical whether the gas is heated at constant volume or constant pressure. That single fact is what turns the definitions above into a clean algebraic relation, worked out next. The ratio of the two,

\gamma = \frac{C_P}{C_V}

is always greater than 1 for an ideal gas. Its numeric value — about $1.67$ for a monatomic gas, $1.4$ for a diatomic gas — comes from the gas's degrees of freedom, a topic developed fully in the Kinetic Theory unit; here $\gamma$ is used only as the ratio $C_P/C_V$ , and it reappears later in this chapter in the adiabatic relation $PV^\gamma = \text{constant}$ .

Derivation AMayer's relation between C sub P and C sub V

The goal is to show, using only the first law of thermodynamics and the ideal gas equation of state for one mole, $PV = RT$ , that the two molar specific heats of an ideal gas differ by exactly the universal gas constant $R$ — no matter what the gas is or what its degrees of freedom are. The entire derivation rests on one physical fact that must be stated before the algebra starts: for an ideal gas, internal energy $U$ is a function of temperature alone, so $\Delta U$ for a given $\Delta T$ is the same number whether volume or pressure was held fixed while that $\Delta T$ occurred.

Step 1. Write the first law for one mole of gas heated at constant volume through $\Delta T$ . Since $\Delta V = 0$ , no work is done, and the first law $\Delta Q = \Delta U + \Delta W$ reduces to $\Delta Q_V = \Delta U$ . By the definition of $C_V$ , $\Delta Q_V = C_V\Delta T$ , so $\Delta U = C_V\Delta T$ . This is the rise in internal energy for a temperature change $\Delta T$ , and — because $U$ depends only on $T$ — it is the same $\Delta U$ that would occur for the same $\Delta T$ under any other condition, including constant pressure.

Step 2. Write the first law for the same one mole heated at constant pressure through the same $\Delta T$ . Now the gas also does work $\Delta W = P\Delta V$ as it expands, so $\Delta Q_P = \Delta U + P\Delta V$ . By the definition of $C_P$ , $\Delta Q_P = C_P\Delta T$ , and from Step 1 the $\Delta U$ here is the same $C_V\Delta T$ . So $C_P\Delta T = C_V\Delta T + P\Delta V$ .

Step 3. Evaluate $P\Delta V$ using the equation of state. For one mole, $PV = RT$ . At constant pressure, differencing this equation gives $P\Delta V = R\Delta T$ (the pressure factors out because it does not change). Substitute this into the equation from Step 2: $C_P\Delta T = C_V\Delta T + R\Delta T$ .

Step 4. Divide through by $\Delta T$ to isolate the relation between the two molar specific heats: $C_P = C_V + R$ , or equivalently

C_P - C_V = R

This is Mayer's relation. It holds for any ideal gas — monatomic, diatomic or polyatomic — because nothing in the derivation used the gas's internal structure; the only physics used was the first law and the ideal-gas equation of state. The relation also shows immediately that $C_P > C_V$ always, since $R > 0$ : constant-pressure heating is unavoidably more expensive by exactly $R$ joules per mole per kelvin, the price of the expansion work.

Worked examples

Board level - C_P and gamma for a monatomic ideal gas from Mayer's relation

For a monatomic ideal gas, $C_V = \dfrac{3}{2}R$ . Using Mayer's relation, find $C_P$ and $\gamma$ . Take $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ .

Step 1. Apply Mayer's relation directly: $C_P = C_V + R = \dfrac{3}{2}R + R = \dfrac{5}{2}R$ .

Step 2. Substitute the numeric value of $R$ : $C_P = \dfrac{5}{2}\times 8.31 = 20.775\ \text{J mol}^{-1}\text{K}^{-1}$ , and $C_V = \dfrac{3}{2}\times 8.31 = 12.465\ \text{J mol}^{-1}\text{K}^{-1}$ .

Step 3. Form the ratio $\gamma = C_P/C_V = \dfrac{5/2}{3/2} = \dfrac{5}{3} \approx 1.67$ . Answer: $C_P = 20.775\ \text{J mol}^{-1}\text{K}^{-1}$ , $C_V = 12.465\ \text{J mol}^{-1}\text{K}^{-1}$ , and $\gamma \approx 1.67$ , matching the known value for monatomic gases such as helium and argon.

Competitive level - heat, internal energy and work for 2 moles heated two ways

2 moles of an ideal gas are heated so the temperature rises from 300 K to 350 K. Given $C_V = 20.8\ \text{J mol}^{-1}\text{K}^{-1}$ , find the heat supplied and the change in internal energy if this is done at constant volume. If the same $\Delta T$ instead occurs at constant pressure with $C_P = 29.1\ \text{J mol}^{-1}\text{K}^{-1}$ , find the heat supplied then, and the extra heat that went into work. Take $R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$ .

Step 1. Constant-volume case: $\Delta Q_V = nC_V\Delta T = 2\times20.8\times50 = 2080\ \text{J}$ . Since $\Delta V = 0$ , no work is done, so all of this heat is internal energy: $\Delta U = \Delta Q_V = 2080\ \text{J}$ .

Step 2. Constant-pressure case, same $\Delta T = 50\ \text{K}$ : $\Delta Q_P = nC_P\Delta T = 2\times29.1\times50 = 2910\ \text{J}$ .

Step 3. Since internal energy depends only on temperature, $\Delta U$ is the same $2080\ \text{J}$ in both cases — this is the key fact from the derivation above applied to a real number. So the extra heat at constant pressure went entirely into work: $\Delta W = \Delta Q_P - \Delta U = 2910 - 2080 = 830\ \text{J}$ .

Step 4. Check independently using Mayer's relation: $n(C_P - C_V)\Delta T = nR\Delta T = 2\times8.3\times50 = 830\ \text{J}$ , matching Step 3 exactly, since this is just $nP\Delta V$ , the total expansion work done by the gas. Answer: at constant volume, $\Delta Q_V = \Delta U = 2080\ \text{J}$ and $\Delta W = 0$ ; at constant pressure, $\Delta Q_P = 2910\ \text{J}$ , $\Delta U = 2080\ \text{J}$ (unchanged), and $\Delta W = 830\ \text{J}$ . Timing note: about 90 seconds. The fast route is $\Delta Q = nC\Delta T$ for each case, then reuse $\Delta U$ unchanged (it depends only on $\Delta T$ ) rather than recomputing it, and get $\Delta W$ by subtraction or by the shortcut $nR\Delta T$ — both must agree, and checking that agreement is the fastest way to catch an arithmetic slip under exam time pressure.

Common traps

Assuming $C_P = C_V$ for gases, a habit carried over from solids and liquids, where the two are nearly equal because volume barely changes on heating. For a gas the gap is exactly $R$ per mole and is never negligible.
Recomputing $\Delta U$ separately for the constant-pressure case instead of reusing the constant-volume value for the same $\Delta T$ — wasted work, and a common source of arithmetic mismatch, since $\Delta U$ depends on $T$ alone for an ideal gas.
Mixing up specific heat capacity per unit mass ( $c$ , units J kg $^{-1}$ K $^{-1}$ ) with molar specific heat ( $C$ , units J mol $^{-1}$ K $^{-1}$ ) — always check which one a question is asking for, and whether the given quantity is mass or number of moles.
Using a numeric value of $\gamma$ (like $1.4$ or $1.67$ ) without knowing where it comes from — Mayer's relation only gives $C_P - C_V = R$ ; the individual values of $C_V$ and $C_P$ (and hence $\gamma$ ) come from the gas's degrees of freedom, covered in the Kinetic Theory unit.
Forgetting the sign or size of $P\Delta V$ when checking work done — for a gas that expands (constant-pressure heating), $\Delta V > 0$ and the gas does positive work on the surroundings, which is why $C_P > C_V$ and not the other way round.

Common trapC_P and C_V look interchangeable but are not — the R-sized gap is real physics, not rounding

The single biggest misconception is treating $C_P$ and $C_V$ as two names for the same number, perhaps rounded differently, because that intuition is correct for solids and liquids. For an ideal gas it is flatly wrong: $C_P - C_V = R \approx 8.3\ \text{J mol}^{-1}\text{K}^{-1}$ is not a small correction, it is often a third or more of $C_V$ itself. The reason is physical, not mathematical — constant-pressure heating of a gas must pay an extra, unavoidable energy cost (the work of pushing back the surroundings as the gas expands) that constant-volume heating never incurs, because a gas locked at fixed volume cannot push anything. Whenever a problem heats a gas and does not say "at constant volume" or "at constant pressure," that condition changes the numeric answer, and assuming the wrong one — or assuming it does not matter — is the single most common way this topic loses marks.

You should be able to…

Define $C_V$ and $C_P$ for an ideal gas and explain, in terms of the first law, why $C_P > C_V$ .
Derive Mayer's relation $C_P - C_V = R$ starting from the first law and $PV = RT$ for one mole.
Calculate $\Delta Q$ , $\Delta W$ and $\Delta U$ for a given number of moles heated through a given $\Delta T$ , at constant volume and at constant pressure.
Use the fact that $\Delta U$ for an ideal gas depends only on $\Delta T$ to shortcut a calculation instead of redoing it for each condition.
State $\gamma = C_P/C_V$ and know that its numeric value comes from the gas's degrees of freedom (Kinetic Theory unit), not from Mayer's relation.
Distinguish molar specific heat (J mol $^{-1}$ K $^{-1}$ ) from specific heat capacity per unit mass (J kg $^{-1}$ K $^{-1}$ ) and convert correctly between mass and moles.

Ask the tutor: Want a walk-through of how the numeric values of $C_V$ and $C_P$ arise from a gas's degrees of freedom, or more mixed practice combining Mayer's relation with constant-volume and constant-pressure heat calculations? Ask the AI tutor for guided examples on gas specific heats.

Why does a gas even need two specific heats when a solid or liquid gets by with just one?

Because solids and liquids barely change volume when heated, so whether you technically hold pressure or volume fixed makes almost no difference — the work term $P\Delta V$ is tiny. A gas, in contrast, expands a lot when heated at constant pressure, so the work it does pushing back the surroundings is a real, sizeable chunk of the heat supplied. That extra cost is exactly what separates $C_P$ from $C_V$ , and it is large enough that ignoring it gives visibly wrong numbers, unlike for solids or liquids where the distinction is safely ignorable.

Is Mayer's relation only true for monatomic gases, or does it work for any gas?

It works for any ideal gas — monatomic, diatomic, or polyatomic. Look back at the derivation: nowhere did it use the number of degrees of freedom or what the gas molecules look like. It only used the first law of thermodynamics and the ideal gas equation $PV=RT$ , both of which hold for every ideal gas. What does depend on the gas's structure is the individual values of $C_V$ and $C_P$ themselves (via degrees of freedom, in the Kinetic Theory unit) — but their difference is always exactly $R$ .

If C_V and C_P are usually written per mole, how do I handle a mass given in kilograms instead of moles?

Convert first. If you are given a mass $m$ and the gas's molar mass $M$ , the number of moles is $n = m/M$ , and you use $\Delta Q = nC\Delta T$ as usual with the molar specific heat $C$ . Alternatively, some problems give a specific heat capacity per unit kilogram, $c$ , directly — in that case use $\Delta Q = mc\Delta T$ and do not multiply by moles at all. The trap is mixing the two conventions in one calculation; always check the units quoted (J mol $^{-1}$ K $^{-1}$ means molar, J kg $^{-1}$ K $^{-1}$ means per unit mass) before plugging numbers in.

Why is gamma always bigger than 1, and does that mean C_V could be negative?

$\gamma = C_P/C_V$ is bigger than 1 simply because $C_P = C_V + R$ and $R$ is a positive constant, so $C_P$ is always the larger of the two — the ratio of a bigger positive number to a smaller positive number is always greater than 1. Both $C_V$ and $C_P$ are themselves always positive for any real gas: physically, it always takes a positive amount of heat to raise a gas's temperature, whether at fixed volume or fixed pressure, so a negative $C_V$ would make no physical sense here.