Question

Atmosphere of Titan. Titan, the largest satellite of Saturn, has a thick nitrogen atmosphere. At its surface, the pressure is 1.5 earth-atmospheres and the temperature is 94 K. (a) What is the surface temperature in °C? (b) Calculate the surface density in Titan’s atmosphere in molecules per cubic meter. (c) Compare the density of Titan’s surface atmosphere to the density of earth’s atmosphere at 22°C. Which body has denser atmosphere?

Short answer

(a) The surface temperature of Titan is -179.15°C.

Step-by-step solution

Step 1

Ideal - gas equation on a molecular basis:

$\mathit{p}\mathit{V}\mathbf{=}\mathit{n}\mathit{K}\mathit{T}$

p = the pressure of the gas, V = the volume of the gas, n = the number of molecules of the gas, T = temperature of gas, k = 1.381 x 10^-23 J/molecule.K = Boltzmann constant.

Step 2:

Change the temperature of Titan from K to °C.

$$\begin{gathered} T_K=T_c+273.15 \\ {T}_c=T_K-273.15 \\ =94k-273.15 \\ =-179.15°C \end{gathered}$$

The surface temperature of Titan is -179.15°C.

Step 3:

The surface density in molecules per cubic meter is given by:

$\mathit{p}\mathbf{=}\frac{\mathbf{N}}{\mathbf{V}}$

From the ideal gas equation we get:

$$\begin{gathered} pV=NkT \\ {\rho }_{Ti\tan }=\frac{N}{V}=\frac{p}{KT} \\ =\frac{1.5\times 101325Pa}{\left(1.381\times {10}^{-23}J/,molecule.K\right)\left(94K\right)} \\ =1.17\times {10}^{26}molecule/m^3 \end{gathered}$$

The surface density in Titan’s atmosphere is $1.17\times {10}^{26}molecule/m^3$

Step 4:

$$\begin{gathered} {\rho }_{Earth}=\frac{N}{V}=\frac{p}{kT} \\ =\frac{101325Pa}{\left(1.381\times {10}^{-23}J/molecule.K\right)\left(273.15\times 22K\right)} \\ =2.49\times {10}^{25}molecules/m^3 \end{gathered}$$

Since${\rho }_{Ti\tan }>{\rho }_{Earth}$ , Titan has more denser atmosphere.