Internal Energy

A2 · 16 min

Every object contains energy stored in its molecules: the energy of their random jiggling and the energy locked in the forces between them. This is internal energy. It explains why heating raises temperature, why melting ice stays at 0 ∘C0\ ^\circ\text{C}, and why the internal energy of an ideal gas depends only on its temperature. The definition is examined word for word in Paper 4, and questions on it are often the first part of a longer first law of thermodynamics question.

What internal energy is

Zoom in on a block of copper. Its atoms are not still: each one vibrates about its position in the lattice, with kinetic energy that changes from moment to moment. The atoms are also held in place by electrical forces from their neighbours, so each has potential energy depending on how far it is from its neighbours. Neither energy is the same for every atom; at any instant some atoms are moving fast, some slowly, some are squeezed close, some stretched apart. The energies are randomly distributed.

Add all of these up and you get the internal energy.

Definition

The internal energy of a system is the sum of a random distribution of kinetic and potential energies of the molecules in the system.

Each phrase matters:

  • Sum: the total over all the molecules, not an average.
  • Random distribution: the energies are shared out randomly among the molecules. This excludes ordered energy such as the kinetic energy of the block flying through the air, or its gravitational potential energy on a shelf. Throwing a block does not change its internal energy.
  • Kinetic energy: from the random translational, rotational and vibrational motion of the molecules.
  • Potential energy: from the intermolecular forces, which depends on the separation of the molecules.

The two parts in solids, liquids and gases

StateMolecular kinetic energyMolecular potential energy
SolidVibration about fixed positionsLarge and negative: strong bonds, molecules close together
LiquidVibration and slow movement past each otherStill large and negative, slightly less so than in the solid
GasFast random translation (and rotation)Close to zero: molecules far apart, forces tiny
Ideal gasAll of the internal energyExactly zero: no intermolecular forces

The potential energy of bonded molecules is negative, for the same reason that gravitational potential energy is negative: zero is defined as infinite separation, and the molecules attract, so energy must be supplied to pull them apart to zero. Breaking bonds (melting, boiling) raises the potential energy towards zero.

Internal energy is a function of state

The state of a system is described by its measurable properties: for a gas, its pressure, volume and temperature (and the amount present). Internal energy is determined by the state of the system. If you know the state, you know the internal energy, however the system got there.

An analogy: your height above sea level depends only on where you are standing, not on the path you climbed. Internal energy depends only on the state, not on whether it was reached by heating, by doing work, or by some combination.

This has two practical consequences:

  • If a system goes through a series of changes and returns to its starting state (a cycle), its internal energy returns to its starting value. The total change in internal energy around any cycle is zero.
  • Different processes between the same two states always give the same ΔU\Delta U, even though the heating and work done may be completely different.

The amounts of heating and work done are not functions of state. A system does not "contain heat" or "contain work". Heating and work are ways of transferring energy; internal energy is what is stored.

Temperature and internal energy

From kinetic theory, the mean translational kinetic energy of a gas molecule is 32kT\tfrac{3}{2}kT: proportional to the thermodynamic temperature. The same is broadly true for solids and liquids: a higher temperature means faster random molecular motion.

Key result
  • A rise in temperature of an object means an increase in the mean kinetic energy of its molecules, and so an increase in its internal energy.
  • At a change of state (melting, boiling) the temperature is constant, so the mean kinetic energy is constant; the energy supplied increases the potential energy of the molecules. The internal energy still increases.

So a rise in temperature always means an increase in internal energy, but an increase in internal energy does not always mean a rise in temperature.

Why this is not circular

Students sometimes find this confusing: temperature is "defined" by thermal equilibrium, yet it also measures molecular kinetic energy. Both are true. Two objects in thermal equilibrium have no net energy transfer between them; kinetic theory shows that this happens when their molecules have the same mean kinetic energy. The thermodynamic temperature scale and the molecular picture agree.

Absolute zero

At absolute zero (0 K0\ \text{K}) the internal energy of a substance is at its minimum. It is not zero: the molecular potential energy is large and negative, and quantum physics shows that even the kinetic energy cannot fall entirely to zero. At A Level, it is enough to say that at absolute zero the internal energy is a minimum, and that for an ideal gas the molecules would have zero kinetic energy.

The internal energy of an ideal gas

An ideal gas has no intermolecular forces, so its molecules have no potential energy. All of its internal energy is the random kinetic energy of the molecules.

For a monatomic ideal gas (helium, neon, argon), the atoms only have translational kinetic energy, 32kT\tfrac{3}{2}kT each on average. For NN atoms, or nn moles:

Key result
U=32NkT=32nRT=32pVU = \tfrac{3}{2}NkT = \tfrac{3}{2}nRT = \tfrac{3}{2}pV

The internal energy of a fixed amount of ideal gas depends only on its thermodynamic temperature: U∝TU \propto T.

Molecules with more than one atom (such as O2\text{O}_2 or N2\text{N}_2) also rotate, so their internal energy is larger than 32nRT\tfrac{3}{2}nRT, but it is still proportional to TT and still independent of volume. Questions that ask you to calculate UU will either specify a monatomic gas or give you the needed values.

y = 12.465 x (300, 0) -- (300, 3739.5) (0, 3739.5) -- (300, 3739.5)

The graph shows U=32nRTU = \tfrac{3}{2}nRT for 1.0 mol1.0\ \text{mol} of a monatomic ideal gas (UU in J, TT in K): a straight line through the origin with gradient 32nR=12.5 J K−1\tfrac{3}{2}nR = 12.5\ \text{J K}^{-1}. At 300 K300\ \text{K}, U=3.74×103 JU = 3.74 \times 10^{3}\ \text{J}.

Important consequences:

  • Isothermal change (constant temperature): ΔU=0\Delta U = 0 for an ideal gas, even if the pressure and volume change a lot.
  • Same temperature change, same ΔU\Delta U: heating a gas from 300 K300\ \text{K} to 400 K400\ \text{K} at constant volume or at constant pressure gives the same ΔU\Delta U. (The energy supplied by heating is different in the two cases, because at constant pressure the gas also does work as it expands. See the first law of thermodynamics.)
  • For real gases, compressing at constant temperature slightly changes the potential energy because the molecules come closer together, but for gases far from condensing this effect is very small.

Internal energy and changes of state

When a substance melts or boils, energy is supplied at constant temperature. From the table, the potential energy of the molecules rises as bonds are broken or weakened, while the mean kinetic energy stays constant.

  • On melting, the molecules stay roughly as close together; only some bonds are broken. The increase in internal energy is mLfmL_f (the volume change and the work done against the atmosphere are negligible).
  • On boiling, the molecules are separated completely. The internal energy increases a great deal, but not by all of mLvmL_v: part of the energy supplied is used to do work pushing back the atmosphere as the vapour expands. The increase in internal energy is mLv−pΔVmL_v - p\Delta V.

This is a first glimpse of the first law: energy supplied by heating goes partly into internal energy and partly into work done by the system.

Explaining internal energy changes
  1. Identify whether the temperature changes. If it rises, the mean molecular kinetic energy increases.
  2. Identify whether the molecular separation changes (change of state, expansion of a real substance). If molecules move apart against attractive forces, the potential energy increases.
  3. Internal energy is the sum of both. State which part changes and in which direction.
  4. For an ideal gas, there is no potential energy: ΔU\Delta U depends only on ΔT\Delta T, and ΔU=32nRΔT\Delta U = \tfrac{3}{2}nR\Delta T for a monatomic gas.

Worked examples

Internal energy of helium

Calculate the internal energy of 2.0 mol2.0\ \text{mol} of helium, treated as a monatomic ideal gas, at 300 K300\ \text{K}, and the increase in internal energy when it is heated to 400 K400\ \text{K}.

SolutionU=32nRT=1.5×2.0×8.31×300=7.48×103 JU = \tfrac{3}{2}nRT = 1.5 \times 2.0 \times 8.31 \times 300 = 7.48 \times 10^{3}\ \text{J}

At 400 K400\ \text{K}: U=1.5×2.0×8.31×400=9.97×103 JU = 1.5 \times 2.0 \times 8.31 \times 400 = 9.97 \times 10^{3}\ \text{J}.

ΔU=32nRΔT=1.5×2.0×8.31×100=2.49×103 J\Delta U = \tfrac{3}{2}nR\Delta T = 1.5 \times 2.0 \times 8.31 \times 100 = 2.49 \times 10^{3}\ \text{J}

This increase is the same however the gas is taken from 300 K300\ \text{K} to 400 K400\ \text{K}.

Internal energy from pressure and volume

A balloon contains helium at a pressure of 1.0×105 Pa1.0 \times 10^{5}\ \text{Pa} in a volume of 0.010 m30.010\ \text{m}^3. Calculate the internal energy of the helium. State one assumption.

Solution

For a monatomic ideal gas, U=32NkTU = \tfrac{3}{2}NkT and pV=NkTpV = NkT, so

U=32pV=1.5×1.0×105×0.010=1.5×103 JU = \tfrac{3}{2}pV = 1.5 \times 1.0 \times 10^{5} \times 0.010 = 1.5 \times 10^{3}\ \text{J}

Assumption: helium behaves as an ideal gas, so its molecules have no potential energy and all its internal energy is translational kinetic energy. The temperature is not needed.

Explaining melting

A block of ice at 0 ∘C0\ ^\circ\text{C} is heated until it has just melted, still at 0 ∘C0\ ^\circ\text{C}. (a) State and explain what happens to the kinetic energy and the potential energy of the molecules. (b) Calculate the increase in internal energy for 0.50 kg0.50\ \text{kg} of ice. (Lf=3.34×105 J kg−1L_f = 3.34 \times 10^{5}\ \text{J kg}^{-1})

Solution

(a) The temperature is constant, so the mean kinetic energy of the molecules is unchanged. The energy supplied breaks some of the intermolecular bonds, so the molecules' potential energy increases (becomes less negative). The internal energy therefore increases.

(b) The volume change on melting is tiny, so almost no work is done on the surroundings; all the energy supplied becomes internal energy:

ΔU=mLf=0.50×3.34×105=1.67×105 J\Delta U = mL_f = 0.50 \times 3.34 \times 10^{5} = 1.67 \times 10^{5}\ \text{J}
Boiling water: where the energy goes

1.0 kg1.0\ \text{kg} of water (M=18 g mol−1M = 18\ \text{g mol}^{-1}) at 100 ∘C100\ ^\circ\text{C} is boiled completely into steam at 100 ∘C100\ ^\circ\text{C} and atmospheric pressure 1.01×105 Pa1.01 \times 10^{5}\ \text{Pa}. The specific latent heat of vaporisation is 2.26×106 J kg−12.26 \times 10^{6}\ \text{J kg}^{-1} and the volume of the liquid water is 1.0×10−3 m31.0 \times 10^{-3}\ \text{m}^3. Treating steam as an ideal gas, calculate (a) the volume of the steam, (b) the work done by the steam in pushing back the atmosphere, and (c) the increase in internal energy.

Solution

(a) n=1000/18=55.6 moln = 1000/18 = 55.6\ \text{mol} and T=373 KT = 373\ \text{K}:

V=nRTp=55.6×8.31×3731.01×105=1.70 m3V = \frac{nRT}{p} = \frac{55.6 \times 8.31 \times 373}{1.01 \times 10^{5}} = 1.70\ \text{m}^3

(b) The pressure is constant, so

W=pΔV=1.01×105×(1.70−0.001)=1.72×105 JW = p\Delta V = 1.01 \times 10^{5} \times (1.70 - 0.001) = 1.72 \times 10^{5}\ \text{J}

(c) The energy supplied by heating is mLv=2.26×106 JmL_v = 2.26 \times 10^{6}\ \text{J}. Of this, 1.72×105 J1.72 \times 10^{5}\ \text{J} leaves the system as work done on the atmosphere. The rest stays as internal energy:

ΔU=2.26×106−1.72×105=2.09×106 J\Delta U = 2.26 \times 10^{6} - 1.72 \times 10^{5} = 2.09 \times 10^{6}\ \text{J}

About 92%92\% of the latent heat goes into increasing the potential energy of the molecules as they are separated; about 8%8\% is the work done against the atmosphere.

Same temperature, different internal energy

Container X holds 1.0 mol1.0\ \text{mol} of argon at 300 K300\ \text{K}. Container Y holds 3.0 mol3.0\ \text{mol} of argon at 200 K200\ \text{K}. Both behave as monatomic ideal gases. (a) Which gas has the greater mean kinetic energy per atom? (b) Which has the greater internal energy? (c) The containers are connected by a thin tube and allowed to reach equilibrium with no energy lost to the surroundings. Calculate the final temperature.

Solution

(a) X: mean kinetic energy per atom is 32kT\tfrac{3}{2}kT, which depends only on temperature, and 300 K>200 K300\ \text{K} > 200\ \text{K}.

(b) UX=1.5×1.0×8.31×300=3.74×103 JU_X = 1.5 \times 1.0 \times 8.31 \times 300 = 3.74 \times 10^{3}\ \text{J}; UY=1.5×3.0×8.31×200=7.48×103 JU_Y = 1.5 \times 3.0 \times 8.31 \times 200 = 7.48 \times 10^{3}\ \text{J}. Y has twice the internal energy despite being colder, because it has three times as many atoms.

(c) No energy is lost and no work is done on the surroundings (the total volume is fixed), so the total internal energy is conserved:

32(1.0+3.0)RT=32R(1.0×300+3.0×200)⇒T=300+6004.0=225 K\tfrac{3}{2}(1.0 + 3.0)RT = \tfrac{3}{2}R(1.0 \times 300 + 3.0 \times 200) \quad\Rightarrow\quad T = \frac{300 + 600}{4.0} = 225\ \text{K}

Temperature tells you the energy per molecule; internal energy is the total.

Watch out

"Internal energy is the kinetic and potential energy of the molecules." This loses the mark: the definition needs "sum of a random distribution of kinetic and potential energies of the molecules". "Random" is what excludes ordered motion of the whole object.

Watch out

Temperature constant means internal energy constant. Only for an ideal gas. During melting or boiling the temperature is constant but the internal energy increases, because the molecular potential energy increases.

Watch out

Giving an ideal gas potential energy. By assumption there are no intermolecular forces in an ideal gas, so there is no molecular potential energy at all. Compressing an ideal gas at constant temperature does not change its internal energy.

Watch out

"Heat energy" or "the heat in a body". Bodies contain internal energy, not heat. Heating (thermal energy transfer) is a process that transfers energy because of a temperature difference. Use the words "internal energy" for what is stored.

Exam tip
  • "Define internal energy" (2 marks): sum of the random distribution; of the kinetic and potential energies of the molecules. Learn this sentence exactly.
  • "Explain why the internal energy of an ideal gas is equal to the total kinetic energy of its molecules" (2 marks): there are no intermolecular forces; so there is no (molecular) potential energy.
  • "State what is meant by a function of state" or "Explain why the change in internal energy around a cycle is zero": internal energy depends only on the state of the system (its pp, VV, TT); after a cycle it is back in its original state.
  • When asked how internal energy changes during a process, refer to both kinetic and potential energies and say which changes: for example "temperature constant so mean kinetic energy constant; molecules separate so potential energy increases".
  • Examiners often contrast a solid being heated (kinetic and potential energy both increase) with a solid melting (only potential energy increases).
Summary
  • Internal energy: the sum of a random distribution of kinetic and potential energies of the molecules in a system.
  • Internal energy is determined by the state of the system; around a complete cycle ΔU=0\Delta U = 0.
  • A rise in temperature means an increase in mean molecular kinetic energy and so an increase in internal energy.
  • At a change of state, temperature (and kinetic energy) is constant; potential energy increases.
  • An ideal gas has no intermolecular forces, so no potential energy; its internal energy is all kinetic and depends only on TT.
  • Monatomic ideal gas: U=32NkT=32nRT=32pVU = \tfrac{3}{2}NkT = \tfrac{3}{2}nRT = \tfrac{3}{2}pV.
  • On boiling, ΔU=mLv−pΔV\Delta U = mL_v - p\Delta V; on melting, ΔU≈mLf\Delta U \approx mL_f.

Practice questions

Question
  1. Define internal energy.
  2. Explain why the internal energy of a stone does not change when it is thrown upwards, even though its kinetic and potential energies change.
  3. Calculate the internal energy of 5.0 mol5.0\ \text{mol} of argon (a monatomic ideal gas) at 20 ∘C20\ ^\circ\text{C}, and the increase in internal energy when it is heated to 80 ∘C80\ ^\circ\text{C}.
  4. A cylinder of volume 0.050 m30.050\ \text{m}^3 contains helium at 2.0×107 Pa2.0 \times 10^{7}\ \text{Pa}. Calculate the internal energy of the helium.
  5. A 0.20 kg0.20\ \text{kg} copper block (specific heat capacity 390 J kg−1 K−1390\ \text{J kg}^{-1}\ \text{K}^{-1}) is heated from 20 ∘C20\ ^\circ\text{C} to 70 ∘C70\ ^\circ\text{C}. Ignoring its tiny expansion, state the increase in its internal energy and describe what happens to its molecules.
  6. Describe the changes in the kinetic energy, potential energy and internal energy of the molecules of water when (a) water is heated from 20 ∘C20\ ^\circ\text{C} to 100 ∘C100\ ^\circ\text{C}, and (b) water at 100 ∘C100\ ^\circ\text{C} boils into steam at 100 ∘C100\ ^\circ\text{C}.
  7. A fixed mass of ideal gas is compressed to half its volume at constant temperature. State and explain the change in its internal energy.
  8. A gas is taken around a cycle A to B to C and back to A. In the stage A to B its internal energy increases by 400 J400\ \text{J}; in B to C it decreases by 150 J150\ \text{J}. State the change in internal energy in C to A, and explain your answer.
  9. 2.0 kg2.0\ \text{kg} of water at 100 ∘C100\ ^\circ\text{C} boils at a constant pressure of 1.01×105 Pa1.01 \times 10^{5}\ \text{Pa} to form steam occupying 3.4 m33.4\ \text{m}^3. The volume of the liquid is negligible. (Lv=2.26×106 J kg−1L_v = 2.26 \times 10^{6}\ \text{J kg}^{-1}) Calculate the work done by the steam on the atmosphere and the increase in internal energy of the water, and explain why the increase in internal energy is less than the energy supplied.
  10. A sealed rigid container holds 0.40 mol0.40\ \text{mol} of neon at 250 K250\ \text{K} and a second sealed rigid container holds 0.60 mol0.60\ \text{mol} of helium at 400 K400\ \text{K}. Both are monatomic ideal gases. They are placed in thermal contact, insulated from the surroundings, until they reach the same temperature. (a) Calculate the final temperature. (b) Calculate the change in internal energy of each gas and comment on the total.
Answers
  1. The sum of a random distribution of kinetic and potential energies of the molecules in a system.
  2. Internal energy only includes the random kinetic and potential energies of the molecules. Throwing the stone gives all its molecules the same ordered motion and raises them together; the random motion and the separations of the molecules (and so the temperature) are unchanged.
  3. T=293 KT = 293\ \text{K}: U=1.5×5.0×8.31×293=1.83×104 JU = 1.5 \times 5.0 \times 8.31 \times 293 = 1.83 \times 10^{4}\ \text{J}. ΔU=1.5×5.0×8.31×60=3.74×103 J\Delta U = 1.5 \times 5.0 \times 8.31 \times 60 = 3.74 \times 10^{3}\ \text{J}.
  4. U=32pV=1.5×2.0×107×0.050=1.5×106 JU = \tfrac{3}{2}pV = 1.5 \times 2.0 \times 10^{7} \times 0.050 = 1.5 \times 10^{6}\ \text{J} (assuming ideal behaviour).
  5. ΔU=mcΔθ=0.20×390×50=3.9×103 J\Delta U = mc\Delta\theta = 0.20 \times 390 \times 50 = 3.9 \times 10^{3}\ \text{J}. The atoms vibrate with larger amplitude and higher mean speed, so their mean kinetic energy increases; their mean separation increases very slightly, so their potential energy increases slightly too.
  6. (a) Temperature rises: mean kinetic energy increases; potential energy increases slightly as the liquid expands; internal energy increases. (b) Temperature constant: mean kinetic energy unchanged; molecules are separated against intermolecular forces, so potential energy increases greatly; internal energy increases.
  7. No change. The internal energy of an ideal gas is all kinetic energy, which depends only on temperature; the temperature is constant. (There are no intermolecular forces, so bringing the molecules closer does not change any potential energy.)
  8. Internal energy is a function of state, so the total change around the cycle is zero: 400−150+ΔUCA=0400 - 150 + \Delta U_{CA} = 0, so ΔUCA=−250 J\Delta U_{CA} = -250\ \text{J} (a decrease of 250 J250\ \text{J}).
  9. W=pΔV=1.01×105×3.4=3.43×105 JW = p\Delta V = 1.01 \times 10^{5} \times 3.4 = 3.43 \times 10^{5}\ \text{J}. Energy supplied =mLv=2.0×2.26×106=4.52×106 J= mL_v = 2.0 \times 2.26 \times 10^{6} = 4.52 \times 10^{6}\ \text{J}. ΔU=4.52×106−3.43×105=4.18×106 J\Delta U = 4.52 \times 10^{6} - 3.43 \times 10^{5} = 4.18 \times 10^{6}\ \text{J}. Some of the energy supplied by heating is transferred out of the system as work done pushing back the atmosphere as the steam expands.
  10. (a) Total internal energy is conserved (rigid containers, so no work; insulated, so no heating from outside): (0.40+0.60)T=0.40×250+0.60×400=340(0.40 + 0.60)T = 0.40 \times 250 + 0.60 \times 400 = 340, so T=340 KT = 340\ \text{K}. (b) Neon: ΔU=1.5×0.40×8.31×90=+449 J\Delta U = 1.5 \times 0.40 \times 8.31 \times 90 = +449\ \text{J}. Helium: ΔU=1.5×0.60×8.31×(−60)=−449 J\Delta U = 1.5 \times 0.60 \times 8.31 \times (-60) = -449\ \text{J}. The total change is zero: energy has been transferred by heating from the helium to the neon.

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