# Relative humidity: seeing your breath

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.43.

The vapour pressure equation gives the phase boundary curve between the liquid and gas phases of a substance if we can assume that the gas is an ideal gas:

$\displaystyle P_{v}=Ke^{-L/RT} \ \ \ \ \ (1)$

where ${K}$ is a constant and ${L}$ is the latent heat of vapourization. We can use this equation to get a rough solution to an everyday situation. When you’re outdoors in cold weather, you can frequently see your breath, which is caused by the humidity in your breath condensing in the colder air because the combination of the vapour pressure in your breath and the outside air exceeds the equilibrium pressure for the ambient temperature.

To get a rough idea how this works, we’ll throw in some numbers. Suppose the breath you exhale has a temperature of ${35^{\circ}\mbox{ C}}$ (normal body temperature is ${37^{\circ}\mbox{C}}$) and 90% humidity, and the outside air has a temperature of ${10^{\circ}\mbox{ C}}$, with an unknown humidity that you wish to find.

To deal with this situation, we can use the value of ${K}$ that we found earlier, which is based on the data in Schroeder’s Figure 5.11 over a temperature range of ${0^{\circ}\mbox{ C}}$ to ${40^{\circ}\mbox{ C}}$. This gives ${K=1.63\times10^{11}\mbox{ Pa}}$. Also, ${L=43.99\times10^{3}\mbox{ J mol}^{-1}}$ and the gas constant is ${R=8.314}$ in SI units.

The equilibrium pressure at ${35^{\circ}\mbox{ C}=308\mbox{ K}}$ is then

 $\displaystyle P_{v}\left(308\right)$ $\displaystyle =$ $\displaystyle \left(1.63\times10^{11}\right)e^{-43.99\times10^{3}/\left(8.314\right)\left(308\right)}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 5643\mbox{ Pa} \ \ \ \ \ (3)$

The vapour pressure in your exhaled breath at 90% humidity is therefore

$\displaystyle P=0.9\times5643=5079\mbox{ Pa} \ \ \ \ \ (4)$

The temperature of the gas just outside your nose after you exhale will be some value between ${35^{\circ}\mbox{ C}}$ and ${10^{\circ}\mbox{ C}}$. Suppose we just take the average (which would occur if you mix equal volumes of the two gases) and say that the temperature of the air just outside your nose is

$\displaystyle T_{nose}=\frac{283+308}{2}=295.5\mbox{ K} \ \ \ \ \ (5)$

The equilibrium vapour pressure at this temperature can be calculated from 1:

$\displaystyle P_{v}\left(295.5\right)=2729\mbox{ Pa} \ \ \ \ \ (6)$

If you can see your breath, the vapour pressure outside your nose must be at least this much. If we’re also assuming that the vapour pressure is the average of the pressure in the air and the vapour pressure that you exhale, then the vapour pressure ${P_{air}}$ of the air must satisfy

$\displaystyle \frac{5079+P_{air}}{2}\ge2729 \ \ \ \ \ (7)$

This gives

$\displaystyle P_{air}\ge378\mbox{ Pa} \ \ \ \ \ (8)$

To calculate the relative humidity of the outside air, we need the equilibrium pressure at ${10^{\circ}\mbox{ C}=283\mbox{ K}}$ which we can again get from 1:

$\displaystyle P_{v}\left(283\right)=1237\mbox{ Pa} \ \ \ \ \ (9)$

Therefore, the humidity of the outside air is at least

$\displaystyle \frac{378}{1237}=0.306 \ \ \ \ \ (10)$

so we’d expect the humidity to be greater than around 30%.

I’m not sure how accurate is the assumption that we can just take the averages of temperature and vapour pressure in the gas into which you exhale, but it shouldn’t be too far off.

One interesting fact comes out of this analysis. It’s not necessary for the air to be really cold in order for you to see your breath. All that’s needed is for the outside air to be just slightly cooler than your breath and have a high enough humidity. You can sometimes see this effect on days when the air is exceptionally humid (for example, during rainy weather) even if it’s not particularly cold.