Specific Heat Capacity

PhysicsKinetic TheoryFor NEET aspirants

The specific heat capacity of a gas is the heat needed to raise the temperature of a unit amount of it by one kelvin, and for a gas its value depends on the process. Kinetic theory with equipartition predicts it for ideal gases: $C_{V} = \frac{f}{2} R$ , $C_{P} = C_{V} + R$ (Mayer's relation) and $γ = 1 + \frac{2}{f}$ . For solids it gives $3 R$ per mole (Dulong-Petit) and for water about $9 R$ . This page covers specific heat capacity of gases, mixtures and solids, and why it changes with temperature: a steady source of JEE Main and NEET questions.

On this page1C depends on the process2Mayer's relation3C from equipartition4Theory vs experiment5Gas mixtures6Solids7Water

Key Formulas - Quick Reference

Molar heat capacity $C = \frac{1}{n} \frac{d Q}{d T}$ ( $J mol^{- 1} K^{- 1}$ ); specific heat capacity $c = \frac{C}{M}$ ( $J kg^{- 1} K^{- 1}$ )
★ Must learnMayer's relation: $C_{P} - C_{V} = R$ per mole; per unit mass $c_{p} - c_{v} = \frac{R}{M}$
★ Must learnFrom equipartition: $C_{V} = \frac{f}{2} R$ , $C_{P} = (\frac{f}{2} + 1) R$ , $γ = \frac{C _{P}}{C _{V}} = 1 + \frac{2}{f}$
★ Must learnMonatomic $\frac{3}{2} R, \frac{5}{2} R, \frac{5}{3}$ ; diatomic $\frac{5}{2} R, \frac{7}{2} R, \frac{7}{5}$ ; non-linear $3 R, 4 R, \frac{4}{3}$
Degrees of freedom from $γ$ : $f = \frac{2}{γ - 1}$ ; also $C_{V} = \frac{R}{γ - 1}$ , $C_{P} = \frac{γ R}{γ - 1}$
★ Must learnMixture: $C_{V, mi x} = \frac{n _{1} C _{V 1} + n _{2} C _{V 2}}{n _{1} + n _{2}}$ , $C_{P, mi x} = C_{V, mi x} + R$
Solids (Dulong-Petit): $C \approx 3 R \approx 25 J mol^{- 1} K^{- 1}$ ; water: $C \approx 9 R \approx 75 J mol^{- 1} K^{- 1}$
Heat supplied: $Q = n C_{V} Δ T$ at constant volume, $Q = n C_{P} Δ T$ at constant pressure

1. Heat Capacity of a Gas Depends on the Process

The heat needed to warm a substance depends on how much of it there is, so heat capacity is quoted per unit mass or per mole:

Specific heat capacity $c = \frac{1}{m} \frac{d Q}{d T}$ ( $J kg^{- 1} K^{- 1}$ ). Molar heat capacity $C = \frac{1}{n} \frac{d Q}{d T}$ ( $J mol^{- 1} K^{- 1}$ ). They are linked by the molar mass: $C = M c$ .

For solids and liquids the volume barely changes on heating, so one value is enough. A gas is different: the heat needed depends on what else happens while it warms. Two processes are standard:

Constant volume ( $C_{V}$ ): the container is rigid, the gas does no work, and all the heat raises the internal energy: $Q = n C_{V} Δ T = Δ U$ .
Constant pressure ( $C_{P}$ ): the gas expands against a constant pressure and does work, so extra heat is needed for the same $Δ T$ : $Q = n C_{P} Δ T = Δ U + P Δ V$ .

Figure 1: For the same

Δ T

the internal energy change is the same (

Δ U = n C_{V} Δ T

). At constant pressure the gas also does work

P Δ V = n R Δ T

lifting the piston, so

Q_{P} = Q_{V} + n R Δ T

and

C_{P} > C_{V}

C_{V}

(constant volume)

Rigid container, $W = 0$ . All heat goes into internal energy: $C_{V} = \frac{1}{n} \frac{d U}{d T}$ . For an ideal gas it fixes $U$ for every process: $Δ U = n C_{V} Δ T$ .

C_{P}

(constant pressure)

Gas expands, does work $P Δ V$ . Heat goes into internal energy and work, so $C_{P} > C_{V}$ . Used for heating in open vessels or under a free piston.

2. Mayer's Relation: $C_{P} - C_{V} = R$

Take $n$ moles of an ideal gas and raise its temperature by $Δ T$ by two different routes (Figure 2).

Figure 2: Both paths start on the isotherm

T

and end on

T + Δ T

, so

Δ U

is the same. Along

A \to B

(constant

V

) no work is done; along

A \to C

(constant

P

) the gas does the shaded work

P Δ V = n R Δ T

. Hence

C_{P} - C_{V} = R

(Mayer's relation).

Constant volume ( $A \to B$ ): $W = 0$ , so $Q_{V} = Δ U = n C_{V} Δ T$ .
Constant pressure ( $A \to C$ ): first law $Q_{P} = Δ U + P Δ V$ , with $Q_{P} = n C_{P} Δ T$ .
For an ideal gas $U$ depends only on $T$ , so $Δ U$ is the same on both routes: $Δ U = n C_{V} Δ T$ .
From $P V = n R T$ at constant $P$ : $P Δ V = n R Δ T$ .
So $n C_{P} Δ T = n C_{V} Δ T + n R Δ T$ . Divide by $n Δ T$ :

★ Must learn

C_{P} - C_{V} = R

The difference is the same for every ideal gas: $8.314 J mol^{- 1} K^{- 1}$ ( $\approx 2 cal mol^{- 1} K^{- 1}$ ). Per unit mass it depends on the gas: $c_{p} - c_{v} = \frac{R}{M}$ .

Exam Trick

Watch the units of " $c_{p} - c_{v}$ ". Per mole it is always $R$ . Per gram it is $\frac{R}{M}$ : for hydrogen $\frac{8.314}{2} = 4.16 J g^{- 1} K^{- 1}$ (about $1 cal g^{- 1} K^{- 1}$ ), for oxygen $\frac{8.314}{32} = 0.26 J g^{- 1} K^{- 1}$ . The lighter the gas, the larger the difference per gram.

3. Heat Capacities from the Law of Equipartition

The previous concept gave the internal energy of an ideal gas with $f$ degrees of freedom: $U = \frac{f}{2} n R T$ . Then

★ Must learn

C_{V} = \frac{1}{n} \frac{d U}{d T} = \frac{f}{2} R, C_{P} = C_{V} + R = (\frac{f}{2} + 1) R, γ = \frac{C _{P}}{C _{V}} = 1 + \frac{2}{f}

The ratio $γ$ (the adiabatic exponent) is fixed by the number of ways a molecule stores energy.

Gas	$f$	$C_{V}$	$C_{P}$	$γ$
Monatomic (He, Ne, Ar)	3	$\frac{3}{2} R = 12.5$	$\frac{5}{2} R = 20.8$	$\frac{5}{3} \approx 1.67$
Diatomic, rigid ( $O_{2}$ , $N_{2}$ )	5	$\frac{5}{2} R = 20.8$	$\frac{7}{2} R = 29.1$	$\frac{7}{5} = 1.40$
Diatomic with vibration	7	$\frac{7}{2} R = 29.1$	$\frac{9}{2} R = 37.4$	$\frac{9}{7} \approx 1.29$
Non-linear, rigid ( $C H_{4}$ , $H_{2} O$ )	6	$3 R = 24.9$	$4 R = 33.3$	$\frac{4}{3} \approx 1.33$

Values in $J mol^{- 1} K^{- 1}$ . A linear polyatomic molecule such as $C O_{2}$ behaves like a diatomic one when rigid.

Figure 3: From equipartition,

C_{V} = \frac{f}{2} R

and

C_{P} = C_{V} + R

. The gap between each pair of bars is always exactly

R

, while

γ = \frac{C _{P}}{C _{V}}

falls as molecules get more complex.

Figure 4:

γ = 1 + \frac{2}{f}

always lies between 1 and

\frac{5}{3}

. The more ways a molecule has to store energy, the closer

C_{P}

and

C_{V}

become in ratio and the closer

γ

gets to 1. Inverting:

f = \frac{2}{γ - 1}

Exam Trick

Everything follows from $γ$ : $f = \frac{2}{γ - 1}$ , $C_{V} = \frac{R}{γ - 1}$ , $C_{P} = \frac{γ R}{γ - 1}$ . So $γ = 1.4$ gives $f = 5$ , $C_{V} = 2.5 R$ ; $γ = \frac{4}{3}$ gives $f = 6$ , $C_{V} = 3 R$ . Remember too: $γ$ always lies between 1 and $\frac{5}{3}$ for an ideal gas.

Quick Recall: tap to check

Find

C_{P}

of a gas with

γ = \frac{5}{3}

C_{P} = \frac{γ R}{γ - 1} = \frac{5}{2} R

(monatomic).

Why is

γ

of argon larger than that of nitrogen?

Argon has fewer degrees of freedom (

3

5

), and

γ = 1 + \frac{2}{f}

C_{P} - C_{V}

larger for

C O_{2}

than for He?

No. Per mole it is

R

for every ideal gas.

4. Theory Compared with Experiment

Measured molar heat capacities near room temperature agree remarkably well with $\frac{f}{2} R$ for monatomic and diatomic gases (Figure 5). For polyatomic gases the measured value is somewhat higher, because some vibrational modes are already partly active.

Figure 5: Equipartition (red lines) fits monatomic and diatomic gases very well. For

C O_{2}

and

C H_{4}

the measured value is higher, because some vibrational modes are already partly active at room temperature.

A deeper problem appears when temperature changes. Classical equipartition says $C_{V}$ should be constant, but it is not. For hydrogen (Figure 6) $C_{V}$ is $\frac{3}{2} R$ below about $80 K$ , rises to $\frac{5}{2} R$ near room temperature, and climbs further only above about $1000 K$ . Rotations and vibrations switch on one after another.

Figure 6:

C_{V}

of hydrogen (curve computed from a simple quantum model with rotational temperature

85 K

and vibrational temperature

6300 K

). Degrees of freedom switch on one after another as

k T

becomes large enough:

\frac{3}{2} R \to \frac{5}{2} R \to

towards

\frac{7}{2} R

. Classical equipartition predicts a constant value; the steps are a quantum effect. Real hydrogen dissociates before reaching

\frac{7}{2} R

JEE Advanced

Why modes freeze. Quantum mechanics allows a rotation or vibration to take energy only in steps. If the step is much larger than $k T$ , collisions can hardly ever excite it and it takes almost no share of energy. For one vibrational mode of frequency $ν$ (Einstein's result), with $x = \frac{h ν}{k T}$ :

C_{v ib} = R \frac{x ^{2} e ^{x}}{( e ^{x} - 1 ) ^{2}}

which tends to $R$ (the classical value) when $k T ≫ h ν$ and to zero when $k T ≪ h ν$ . For $N_{2}$ , $\frac{h ν}{k} \approx 3400 K$ , so at $300 K$ the vibrational contribution is under $0.01 R$ . The same idea explains the fall of $C$ of solids at low temperature, where $C \propto T^{3}$ (Debye law).

Key idea

Equipartition gives the right

C_{V}

only for the degrees of freedom that are actually active at that temperature. At room temperature: translation and rotation yes, vibration mostly no.

5. Heat Capacity of a Gas Mixture

When gases are mixed at the same temperature, their internal energies add: $U = n_{1} C_{V 1} T + n_{2} C_{V 2} T$ . So the mixture behaves like a single gas with a mole-weighted heat capacity:

Figure 7: Heat capacities of a mixture add because energies add:

C_{V, mi x} = \frac{n _{1} C _{V 1} + n _{2} C _{V 2}}{n _{1} + n _{2}}

. Then

C_{P, mi x} = C_{V, mi x} + R

and

γ_{mi x} = \frac{C _{P, mi x}}{C _{V, mi x}}

. Never average the

γ

values directly.

★ Must learn

C_{V, mi x} = \frac{n _{1} C _{V 1} + n _{2} C _{V 2}}{n _{1} + n _{2}}, C_{P, mi x} = C_{V, mi x} + R, γ_{mi x} = \frac{C _{P, mi x}}{C _{V, mi x}}

Equivalent form: $\frac{n _{1} + n _{2}}{γ _{mi x} - 1} = \frac{n _{1}}{γ _{1} - 1} + \frac{n _{2}}{γ _{2} - 1}$ . Never average the $γ$ values directly.

Exam Trick

Equal moles of a monatomic and a diatomic gas give $γ_{mi x} = 1.5$ . $C_{V, mi x} = \frac{1.5 R + 2.5 R}{2} = 2 R$ , $C_{P, mi x} = 3 R$ , $γ = \frac{3}{2}$ . The simple average of $1.67$ and $1.40$ would give $1.53$ : close, but it is the wrong method and fails for other ratios.

6. Specific Heat Capacity of Solids

In a solid each atom is held at its lattice site by forces from its neighbours, which act like springs. An atom cannot wander, but it oscillates about its mean position in three independent directions.

Figure 8: In a solid each atom oscillates about its lattice site in three independent directions. Each direction has a kinetic and a potential term, so an atom has

6 \times \frac{1}{2} k T = 3 k T

. For one mole

U = 3 R T

and

C = 3 R \approx 24.9 J mol^{- 1} K^{- 1}

(law of Dulong and Petit). For a solid

Δ V

is tiny, so

C_{P} \approx C_{V}

Each oscillation direction has a kinetic term and a potential term, each getting $\frac{1}{2} k T$ : that is $k T$ per direction.
Three directions: average energy per atom $= 3 k T$ .
For one mole of atoms, $U = 3 N_{A} k T = 3 R T$ .
Heating a solid barely changes its volume, so $Δ Q \approx Δ U$ and $C_{P} \approx C_{V}$ :

★ Must learn

C = \frac{d U}{d T} = 3 R \approx 24.9 J mol^{- 1} K^{- 1}

This is the law of Dulong and Petit: the molar heat capacity of most solid elements at room temperature is about $3 R$ , whatever the element. The specific heat per kilogram is therefore roughly $\frac{3 R}{M}$ : large for light elements, small for heavy ones.

Figure 9: Most solids at room temperature have

C \approx 3 R

(dashed line). Carbon is the striking exception: its light atoms and very stiff bonds mean the vibrations are still mostly frozen at

300 K

The law fails at low temperatures, where all solids have $C \to 0$ , and for light, hard solids such as carbon (diamond), boron and silicon even at room temperature. Their vibrations are still partly frozen, just like the vibrations of gas molecules.

Figure 10: Heat capacity of solids falls to zero at low temperature (curves from the Debye model; Debye temperatures lead

105 K

, copper

343 K

, diamond

2230 K

C

approaches the classical value

3 R

only once

T

is about equal to or above this temperature: at

300 K

copper is already at

23.4

, but the diamond curve gives only

4.1 J mol^{- 1} K^{- 1}

(measured

6.1

; the one-parameter model is rough, but the verdict is clear: far below

3 R

7. Specific Heat Capacity of Water

Liquid water can be treated roughly like a solid made of atoms. A water molecule has three atoms (two hydrogen and one oxygen), each oscillating with energy $3 k T$ :

U = 3 \times 3 k T \times N_{A} = 9 R T \Rightarrow C \approx 9 R \approx 75 J mol^{- 1} K^{- 1}

The measured value is $75.3 J mol^{- 1} K^{- 1}$ , which is $\frac{75.3}{0.018} \approx 4180 J kg^{- 1} K^{- 1}$ , the familiar specific heat of water. The close agreement is partly lucky (hydrogen-bond motions contribute too), but it shows why water, with three atoms per molecule and a small molar mass, has such a large specific heat per kilogram.

Solid element

1 atom per formula unit, 6 quadratic terms per atom: $C \approx 3 R \approx 25 J mol^{- 1} K^{- 1}$ .

Water

3 atoms per molecule, each like an atom in a solid: $C \approx 9 R \approx 75 J mol^{- 1} K^{- 1}$ , i.e. about $4.2 kJ kg^{- 1} K^{- 1}$ .

Quick Recall: tap to check

Why is the molar heat capacity of a solid 3R?

Each atom has 3 oscillation directions with KE and PE terms:

6 \times \frac{1}{2} k T = 3 k T

per atom,

3 R T

per mole.

Why is

C_{P} \approx C_{V}

for a solid?

Its volume hardly changes on heating, so almost no work is done.

Why does diamond disobey the Dulong-Petit law at room temperature?

Its light atoms and stiff bonds give high vibration frequencies, which are still frozen at 300 K.

8. Solving Problems and Revision Map

Most heat capacity problems follow the same four steps.

Figure 11: Heat capacity problems in four steps: identify

f

, get

C_{V}

and

C_{P}

, handle mixtures by mole-weighting

C_{V}

, then use the heat equation for the process.

Figure 12: Mind map of this concept. Recall each branch before you check it.

9. Solved Examples

Solved Example 1

Using equipartition, find

C_{V}

C_{P}

(in

J mol^{- 1} K^{- 1}

) and

γ

for (a) helium, (b) nitrogen, (c) methane, treating the molecules as rigid.

Solution:

(a) $f = 3$ : $C_{V} = 1.5 \times 8.314 = 12.47$ , $C_{P} = 20.79$ , $γ = \frac{5}{3} = 1.67$ .

(b) $f = 5$ : $C_{V} = 2.5 \times 8.314 = 20.79$ , $C_{P} = 29.10$ , $γ = 1.40$ .

Answer: He $12.5, 20.8, 1.67$ ; $N_{2}$ $20.8, 29.1, 1.40$ ; $C H_{4}$ $24.9, 33.3, 1.33$ .

Solved Example 2

1 mol

of helium is mixed with

1 mol

of oxygen (rigid). Find

C_{V}

C_{P}

and

γ

of the mixture.

Solution:

$C_{V, mi x} = \frac{1 ( 1.5 R ) + 1 ( 2.5 R )}{2} = 2 R$ ; $C_{P, mi x} = 2 R + R = 3 R$ .

Answer: $C_{V} = 2 R \approx 16.6$ , $C_{P} = 3 R \approx 24.9 J mol^{- 1} K^{- 1}$ , $γ = 1.5$ .

Solved Example 3

2 mol

of nitrogen are heated from

300 K

350 K

. Find the heat required (a) at constant volume, (b) at constant pressure, and (c) the work done by the gas in (b).

Solution:

(a) $Q_{V} = n C_{V} Δ T = 2 \times \frac{5}{2} (8.314) \times 50 = 2078.5 J$ .

(b) $Q_{P} = n C_{P} Δ T = 2 \times \frac{7}{2} (8.314) \times 50 = 2909.9 J$ .

Answer: (a) $\approx 2.08 kJ$ ; (b) $\approx 2.91 kJ$ ; (c) $\approx 831 J$ .

Solved Example 4

For a gas the ratio of specific heats is

γ = 1.4

. Find the number of degrees of freedom and

C_{V}

Solution:

$f = \frac{2}{γ - 1} = \frac{2}{0.4} = 5$ ; $C_{V} = \frac{R}{γ - 1} = \frac{8.314}{0.4}$ .

Answer: $f = 5$ (rigid diatomic); $C_{V} \approx 20.8 J mol^{- 1} K^{- 1}$ .

Solved Example 5

Find the specific heat capacities

c_{v}

and

c_{p}

of argon in

J kg^{- 1} K^{- 1}

. (Molar mass

40 g mol^{- 1}

)

Solution:

$c_{v} = \frac{C _{V}}{M} = \frac{1.5 \times 8.314}{0.040} = 311.8$ ; $c_{p} = \frac{2.5 \times 8.314}{0.040} = 519.6$ .

Answer: $c_{v} \approx 312$ , $c_{p} \approx 520 J kg^{- 1} K^{- 1}$ ; check: $c_{p} - c_{v} = \frac{R}{M} = 208$ .

Solved Example 6

Estimate the specific heat capacity of aluminium (molar mass

27 g mol^{- 1}

) from the law of Dulong and Petit and compare it with the measured

900 J kg^{- 1} K^{- 1}

Solution:

$c = \frac{3 R}{M} = \frac{3 \times 8.314}{0.027}$ .

Answer: $c \approx 924 J kg^{- 1} K^{- 1}$ , within about $3%$ of the measured value.

Solved Example 7

Treating each atom of a water molecule like an atom in a solid, estimate the molar and the specific heat capacity of water.

Solution:

3 atoms, each with energy $3 k T$ : $U = 9 R T$ per mole, so $C = 9 R = 9 \times 8.314 = 74.8 J mol^{- 1} K^{- 1}$ .

Per kilogram: $c = \frac{74.8}{0.018}$ .

Answer: $C \approx 75 J mol^{- 1} K^{- 1}$ , $c \approx 4.16 \times 1 0^{3} J kg^{- 1} K^{- 1}$ (measured $4186$ ).

Solved Example 8

For hydrogen gas,

c_{p} - c_{v}

per gram is
(A)

8.314 J g^{- 1} K^{- 1}

(B)

4.157 J g^{- 1} K^{- 1}

(C)

16.6 J g^{- 1} K^{- 1}

(D)

2 J g^{- 1} K^{- 1}

Solution:

Answer: (B). Per mole the difference is $R$ ; per gram it is $\frac{R}{M} = \frac{8.314}{2} = 4.157 J g^{- 1} K^{- 1}$ .

Solved Example 9

At a very high temperature the molecules of a diatomic gas vibrate. The value of

γ

is
(A)

\frac{5}{3}

(B)

\frac{7}{5}

(C)

\frac{9}{7}

(D)

\frac{4}{3}

Solution:

Answer: (C). With vibration $f = 7$ : $C_{V} = \frac{7}{2} R$ , $C_{P} = \frac{9}{2} R$ , $γ = \frac{9}{7} \approx 1.29$ .

Solved Example 10

2 mol

of a monatomic gas are mixed with

3 mol

of a diatomic gas (rigid). Find

γ

of the mixture.

Solution:

$C_{V, mi x} = \frac{2 ( 1.5 R ) + 3 ( 2.5 R )}{5} = \frac{10.5 R}{5} = 2.1 R$ ; $C_{P, mi x} = 3.1 R$ .

Check with the other form: $\frac{5}{γ - 1} = \frac{2}{2/3} + \frac{3}{2/5} = 3 + 7.5 = 10.5$ , so $γ - 1 = \frac{5}{10.5}$ .

Answer: $γ_{mi x} = \frac{31}{21} \approx 1.48$ .

Practice Questions

Find $C_{V}$ , $C_{P}$ and $γ$ of rigid methane in terms of $R$ .Answer: $3 R$ , $4 R$ , $\frac{4}{3}$
What is $γ$ for a gas with 7 degrees of freedom?Answer: $\frac{9}{7} \approx 1.29$
How much heat raises the temperature of $4 g$ of helium by $10 K$ at constant pressure?Answer: $\approx 208 J$
Find $γ$ for a mixture of $1 mol$ of argon and $1 mol$ of nitrogen.Answer: $1.5$
Use the Dulong-Petit law to estimate the specific heat of copper ( $M = 63.5 g mol^{- 1}$ ).Answer: $\approx 393 J kg^{- 1} K^{- 1}$ (measured $386$ )
A gas has $γ = \frac{4}{3}$ . How many degrees of freedom does it have?Answer: 6
Find $c_{p} - c_{v}$ for oxygen in $J kg^{- 1} K^{- 1}$ .Answer: $\approx 260 J kg^{- 1} K^{- 1}$

Common Mistakes to Avoid

Watch out

Writing $c_{p} - c_{v} = R$ for specific heats per kilogram or per gram. Per mole it is $R$ ; per unit mass it is $\frac{R}{M}$ .
Averaging $γ$ values for a mixture. Average $C_{V}$ weighted by moles, add $R$ for $C_{P}$ , then divide.
Using $C_{P}$ for heating in a closed rigid vessel. A rigid vessel means constant volume: $Q = n C_{V} Δ T$ .
Forgetting that $Δ U = n C_{V} Δ T$ holds for every process of an ideal gas, not only constant-volume ones.
Taking $γ = 1.4$ for every gas. It is $\frac{5}{3}$ for monatomic, $1.4$ for rigid diatomic and $\frac{4}{3}$ for non-linear gases.
Treating $C O_{2}$ as non-linear. It is linear: rigid $γ = 1.4$ , like a diatomic gas.
Applying Dulong-Petit to every solid at every temperature. It fails at low temperature and for light, hard solids such as diamond.
Counting a vibration as one degree of freedom when finding $C_{V}$ at high temperature. Each active mode adds $R$ to $C_{V}$ , not $\frac{R}{2}$ .

Frequently Asked Questions

What is Mayer's relation?

Mayer's relation states that for an ideal gas the molar heat capacity at constant pressure exceeds that at constant volume by the gas constant: C P minus C V equals R. The extra heat at constant pressure is the work n R delta T done as the gas expands.

How does equipartition give the specific heats of gases?

Equipartition gives the internal energy f over 2 n R T. Differentiating with respect to temperature gives C V equals f over 2 R, and Mayer's relation gives C P equals C V plus R. So a monatomic gas has 1.5 R and 2.5 R, a rigid diatomic gas 2.5 R and 3.5 R.

What is gamma for monatomic, diatomic and polyatomic gases?

Gamma equals 1 plus 2 over f. It is 5 over 3 for monatomic gases, 7 over 5 for rigid diatomic and linear gases, 4 over 3 for rigid non-linear polyatomic gases, and 9 over 7 for a diatomic gas whose vibration is active.

How do you find the specific heat of a mixture of gases?

Take the mole-weighted average of the molar heat capacities at constant volume, then add R to get the value at constant pressure. Gamma of the mixture is the ratio of the two. Averaging the gamma values directly gives the wrong answer.

What is the law of Dulong and Petit?

It states that most solid elements have a molar heat capacity close to 3 R, about 25 joules per mole per kelvin, at room temperature. Each atom oscillates in three directions with kinetic and potential energy, giving 3 k T per atom.

Why do heat capacities decrease at low temperatures?

Rotations and vibrations need a minimum energy step to be excited, a quantum effect. When k T is much smaller than that step, the modes are frozen and take no energy. So hydrogen behaves as monatomic below about 80 kelvin and solids have heat capacities approaching zero.

How are specific heats of gases asked in NEET?

NEET asks for C P, C V or gamma of monatomic and diatomic gases, gamma of a mixture, the degrees of freedom from gamma, and C P minus C V. Learn the values 3 by 2 R, 5 by 2 R, 5 by 3 and 7 by 5 by heart.

What kind of specific heat capacity questions come in JEE Main?

JEE Main asks for gamma and heat capacities of gas mixtures, heat supplied in isobaric and isochoric processes, per-gram differences R over M, Dulong-Petit estimates and the effect of vibrational modes on gamma at high temperature.

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Practice questions →Take a mock test

1. Heat Capacity of a Gas Depends on the Process

2. Mayer's Relation: CP​−CV​=R

3. Heat Capacities from the Law of Equipartition

4. Theory Compared with Experiment

5. Heat Capacity of a Gas Mixture

6. Specific Heat Capacity of Solids

7. Specific Heat Capacity of Water

8. Solving Problems and Revision Map

9. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is Mayer's relation?

How does equipartition give the specific heats of gases?

What is gamma for monatomic, diatomic and polyatomic gases?

How do you find the specific heat of a mixture of gases?

What is the law of Dulong and Petit?

Why do heat capacities decrease at low temperatures?

How are specific heats of gases asked in NEET?

What kind of specific heat capacity questions come in JEE Main?

Ready to master Kinetic Theory?

2. Mayer's Relation: $C_{P} - C_{V} = R$