# 1.6: Equation of State-Ideal Gas Law

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- 9531

Because pressure is caused by the movement of molecules, you might expect the pressure P to be greater where there are more molecules (i.e., greater density ρ), and where they are moving faster (i.e., greater temperature T). The relationship between pressure, density, and temperature is called the Equation of State.

Different fluids have different equations of state, depending on their molecular properties. The gases in the atmosphere have a simple equation of state known as the Ideal Gas Law.

For dry air (namely, air with the usual mix of gases, except no water vapor), the ideal gas law is:

\(\ \begin{align}P=\rho \cdot \Re_{d} \cdot T\tag{1.18}\end{align}\)

where ℜ_{d} = 0.287053 kPa·K^{–1}·m^{3}·kg^{–1} = 287.053 J·K^{–1}·kg^{–1} .

ℜ_{d} is called the gas constant for dry air. Absolute temperatures (K) must be used in the ideal gas law. The total air pressure P is the sum of the partial pressures of nitrogen, oxygen, water vapor, and the other gases.

A similar equation of state can be written for just the water vapor in air:

\(\ \begin{align}e=\rho_{v} \cdot \Re_{v} \cdot T\tag{1.19}\end{align}\)

where e is the partial pressure due to water vapor (called the vapor pressure), ρ_{v} is the density of water vapor (called the absolute humidity), and the gas constant for pure water vapor is

ℜ_{v} = 0.4615 kPa·K^{–1}·m^{3}·kg^{–1} = 461.5 J·K^{–1}·kg^{–1} .

For moist air (normal gases with some water vapor),

\(\begin{align}P=\rho \cdot \Re \cdot T\tag{1.21}\end{align}\)

where density ρ is now the total density of the air. A difficulty with this last equation is that the “gas constant” is NOT constant. It changes as the humidity changes because water vapor has different molecular properties than dry air.

To simplify things, a virtual temperature T_{v} can be defined to include the effects of water vapor:

\(\ \begin{align}T_{v}=T \cdot[1+(a \cdot r)]\tag{1.21}\end{align}\)

where r is the water-vapor mixing ratio [r = (mass of water vapor)/(mass of dry air), with units g_{water}_{ vapor} /g_{dry}_{ air}, see the Water Vapor chapter], a = 0.61 g_{dry}_{ air}/g_{water}_{ vapor}, and all temperatures are in absolute units (K). In a nutshell, moist air of temperature T behaves as dry air with temperature T_{v} . T_{v} is greater than T because water vapor is less dense than dry air, and thus moist air acts like warmer dry air.

If there is also liquid water or ice in the air, then this virtual temperature must be modified to include the liquid-water loading (i.e., the weight of the drops falling at their terminal velocity) and ice loading:

\(\ \begin{align}T_{v}=T \cdot\left[1+(a \cdot r)-r_{L}-r_{I}\right]\tag{1.22}\end{align}\)

where r_{L} is the liquid-water mixing ratio (g_{liquid}_{ water} / g_{dry}_{ air}), r_{I} is the ice mixing ratio (g_{ice} / g_{dry}_{ air}), and a = 0.61 (g_{dry}_{ air} / g _{water vapor} ). Because liquid water and ice are heavy, air with liquid-water and/ or ice loading acts like colder dry air.

With these definitions, a more useful form of the ideal gas law can be written for air of any humidity:

\(\ \begin{align}P=\rho \cdot \Re_{d} \cdot T_{v}\tag{1.23}\end{align}\)

where ℜ_{d} is still the gas constant for dry air. In this form of the ideal gas law, the effects of variable humidity are hidden in the virtual temperature factor, which allows the dry “gas constant” to be used (nice, because it really is constant).

**Sample Application **

What is the average (standard) surface temperature for dry air, given standard pressure and density?

**Find the Answer: **

Given: P = 101.325 kPa, ρ = 1.225 kg·m^{-3}

Find: T = ? K

Solving eq. (1.18) for T gives: T = P / (ρ·ℜ_{d})

\(\ T=\frac{101.325\ kPa}{1.225\ kg\cdot m^{-3})\cdot (0.287\ kPa\cdot K^{-1} \cdot m^3 \cdot kg^{-1})}=288.2\ K\) = **15**^{o}C

**Check:** Units OK. Physically reasonable.

**Exposition:** The answer agrees with the standard surface temperature of 15°C discussed earlier, a cool but pleasant temperature

**Sample Application**

What is the absolute humidity of air of temperature 20°C and water vapor pressure of 2 kPa?

**Find the Answer: **

Given: e = 2 kPa, T = 20°C = 293 K

Find: ρ_{v} = ? kg_{water vapor} ·m^{-3}

Solving eq. (1.19) for ρ_{v} gives: ρv = e / (ℜ_{v}·T)

ρ_{v} = ( 2 kPa ) / ( 0.4615 kPa·K^{–1}·m^{3}·kg^{–1} · 293 K )

= __ 0.0148__ kg

_{water vapor}·m

^{-3 }

**Check:** Units OK. Physically reasonable.

**Exposition:** Small compared to the total air density.

**Sample Application**

In an unsaturated tropical environment with temperature of 35°C and water-vapor mixing ratio of 30 g_{water vapor}/kg_{dry air}, what is the virtual temperature?

**Find the Answer:**

Given: T = 35°C, r = 30 g_{water vapor}/kg_{dry air}

Find: T_{v} = ? °C

First, convert T and r to proper units

T = 273.15 + 35 = 308.15 K.

r =(30 g_{water}/kg_{ air})·(0.001 kg/g) = 0.03 g_{water}/g_{air}

Next use eq. (1.21):

T_{v} = (308.15 K)·[ 1 + (0.61 · 0.03) ]

= 313.6 K = **40.6****°C.**

**Check:** Units OK. Physically reasonable

**Exposition:** Thus, high humidity reduces the density of the air so much that it acts like dry air that is 5°C warmer, for this case.

**Sample Application **

In a tropical environment with temperature of 35°C, water-vapor mixing ratio of 30 g_{water vapor}/kg_{dry air} , and 10 g_{liquid water}/kg_{dry air} of raindrops falling at their terminal velocity through the air, what is the virtual temperature?

**Find the Answer:**

Given: T = 35°C, r = 30 g_{water vapor}/kg_{dry air}

r_{L} = 10 g_{liquid water}/kg_{dry air}

Find: T_{v} = ? °C

First, convert T , r and r_{L} to proper units

T = 273.15 + 35 = 308.15 K.

r =(30 g_{vapor}/kg _{air})·(0.001 kg/g) = 0.03 g_{vapor}/g _{air}

r_{L} =(10 g_{liquid}/kg _{air})·(0.001 kg/g) = 0.01 g_{liquid}/g _{air}

Next use eq. (1.22):

T_{v} = (308.15 K)·[ 1 + (0.61 · 0.03) – 0.01 ] = 310.7 K =__ ____37.6__°C.

**Check:** Units OK. Physically reasonable.

**Exposition:** Compared to the previous Sample Application, the additional weight due to falling rain made the air act like it was about 3°C cooler.