# Boiling of water does not involve temperature change, so why is there a change in internal energy during the process? [duplicate]

in an isothermal process, the value ΔT = 0 and therefore ΔU = 0 but during boiling if water at constant atmospheric pressure temperature doesn't change so why internal energy changes

• The issue here is that this is a process that is isothermal, but involves a phase change. Heat is transferred and work is done, so $\Delta U \ne 0$. See my answer for more detail. – Ben Norris Dec 21 '15 at 12:02

Change in internal energy in isothermal process is zero when no phase change or chemical changes are involved (and then only when changes in pressure are small). Since in boiling the phase changes from liquid to gas, there is also change in internal energy. The energy is required to break bonds between water molecules and is regained when the gas condenses back to liquid.

• Edited. Maybe it's now more accurate even though not complete. – KLuuppo Dec 21 '15 at 11:15
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The change in internal energy $U$ is dependent on the heat transferred $q$ and the work done $w$.

$$\Delta U = q + w$$

$\Delta U = 0$ in two cases: when $q=-w$ and when $q=w=0$.

The heat transferred is not necessarily zero if there is no temperature change. The most common expression of heat transfer is $q=C\Delta T$, where $C$ is the heat capacity of the system and $\Delta T$ is a temperature change. We can also write $q=ms\Delta T$, where $m$ is the mass of the system and $s$ is its specific heat capacity (per unit mass). However, heat is being transferred during boiling (else how does the liquid boil?). We need a new equation for heat that does not require temperature.

In a constant pressure system (most boiling happens in a constant pressure, open to the atmosphere, system), change in enthalpy is equal to $q$ since all of the work and other pressure-volume terms cancel. The enthalpy change of vaporization is usually given in $\mathrm{kJ}$ per mole or unit mass. Now we have a new equation for heat:

$$q_{vap}=m\Delta_{vap}H$$

Let's say we are boiling 100.0 g of water. Water's enthalpy of vaoprization is $2257\ \mathrm{kJ/kg}$. Thus the heat transfer is:

$$q=(0.1000\ \mathrm{kg})(2257\ \mathrm{kJ/kg})=225.7\ \mathrm{kJ}$$ Note that in this case, the work done is also not zero. There is a volume change of the system. Let's take our 100 grams of water.

$$V=0.1000\ \mathrm{kg}\times \dfrac{\mathrm{m^3}}{958.4\ \mathrm{kg}}=1.040\times 10^{-4}\ \mathrm{m^3}$$

At the boiling point, the density is $958.4\ \mathrm{kg/m^3}=0.9584\ \mathrm{kg/L}$. Using the ideal gas law (water vapor is reasonably ideal at atmospheric pressure and its boiling point), we can determine the volume of 100.0 g of water vapor, given that

$$100.0\ \mathrm{g}\ \ce{H2O}\times \dfrac{1\ \mathrm{mol}\ \ce{H2O}}{18.02\ \mathrm{g}\ \ce{H2O}}=5.549\ \mathrm{mol}\ \ce{H2O}$$

$$V=\dfrac{nRT}{P}=\dfrac{(5.549\ \mathrm{mol})(0.080257\ \mathrm{\frac{L\cdot atm}{K\cdot mol}})(373.15\ \mathrm{K})}{1\ \mathrm{atm}}=169.9\ \mathrm{L}$$

We can now calculate work. So that we get joules out, I am going to convert pressure to pascals and volume to $\mathrm{m^3}$. $$P=1\ \mathrm{atm}=101,325\ \mathrm{Pa}\\ V_{gas} = 169.9\mathrm{\ L} =0.1699\ \mathrm{m^3}\\ \Delta V= 0.1699\ \mathrm{m^3} - 1.040\times 10^{-4}\ \mathrm{m^3} =0.1698\ \mathrm{m^3}\\ w=-P\Delta V=-101325\ \mathrm{Pa}\times 0.1698\ \mathrm{m^3}=-17.20\ \mathrm{kJ}$$

Now we calculated $\Delta U$:

$$\Delta U = q +w = 225.7\ \mathrm{kJ} - 17.20\ \mathrm{kJ} = 208.5\ \mathrm{kJ}$$