Chapter 14, Problem 115QAP

To determine

(a)

The density of particles in the atmosphere of Titan.

Answer to Problem 115QAP

The density of particles in the atmosphere of Titan is $1.2\times {10}^{26}{\text{particles/m}}^{\text{3}}$.

Explanation of Solution

Given info:

Temperature at Titan is, $T_1=-179°\text{C}$.

Pressure at Titan is, $P_1=1.5\text{atm}$.

Mass of nitrogen molecule is, $m_{\text{N}}=4.7\times {10}^{-26}\text{kg}$.

Diameter of the sphere is, $d=2.4\times {10}^{-10}\text{m}$.

Formula used:

Formula for the ideal gas equation is,

  $PV=NkT$

Calculation:

The density of particles can be calculated as,

  $\begin{array}{c}\frac{N_1}{V}=\frac{P_1}{k{T}_1}\\ {\rho }_1=\frac{1.5\times 1.03\times {10}^5\text{Pa}}{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(-179+273\right)\text{K}}\\ {\rho }_1=1.2\times {10}^{26}{\text{particles/m}}^{\text{3}}\end{array}$

Conclusion:

Thus, the density of particles in the atmosphere of Titan is $1.2\times {10}^{26}{\text{particles/m}}^{\text{3}}$.

To determine

(b)

The body that have denser atmosphere, Titan or Earth.

Answer to Problem 115QAP

Titan has denser atmosphere.

Explanation of Solution

Given info:

Temperature at Titan is, $T_1=-179°\text{C}$.

Pressure at Titan is, $P_1=1.5\text{atm}$.

Mass of nitrogen molecule is, $m_{\text{N}}=4.7\times {10}^{-26}\text{kg}$.

Diameter of the sphere is, $d=2.4\times {10}^{-10}\text{m}$.

Temperature at Earth is, $T_2=10°\text{C}$.

Pressure at Earth is, $P_2=1\text{atm}$.

Formula used:

Formula for the ideal gas equation is,

  $PV=NkT$

Calculation:

The particle density on Titan can be calculated as,

  ${\rho }_1=\frac{P_1}{k{T}_1}$  ...... (1)

The particle density on Earth can be calculated as,

  ${\rho }_2=\frac{P_2}{k{T}_2}$  ...... (2)

Dividing equation (1) by equation (2), we get

  $\begin{array}{c}\frac{{\rho }_1}{{\rho }_2}=\frac{P_1\times T_2}{P_2\times T_1}\\ \frac{{\rho }_1}{{\rho }_2}=\frac{1.5\times \left(10+273\right)}{1.0\times \left(-179+273\right)}\\ \frac{{\rho }_1}{{\rho }_2}=4.5\\ {\rho }_1=4.5{\rho }_2\end{array}$

So, the particle density on Titan is $4.5$times more than the particle density on Earth. Hence, Titan has denser atmosphere.

Conclusion:

Thus, Titan has denser atmosphere.

To determine

(c)

The average distance that a nitrogen molecule travels between collisions on Titan and compare it with the distance for oxygen.

Answer to Problem 115QAP

The average distance that a nitrogen molecule travels between collisions on Titan is $3.3\times {10}^{-8}\text{m}$and it is about $1.7$times less than the average distance for oxygen. This result is unreasonable.

Explanation of Solution

Given info:

Diameter of the sphere is, $d=2.4\times {10}^{-10}\text{m}$.

Density of particle at Titan is, $\frac{N}{V}=1.2\times {10}^{26}{\text{particles/m}}^{\text{3}}$

Formula used:

The formula for the mean free path is given as,

  $\lambda =\frac{1}{4\sqrt{2}\pi r^2\left(\frac{N}{V}\right)}$

Calculation:

The average distance that a nitrogen molecule travels between collisions can be calculated as,

  $\begin{array}{c}\lambda =\frac{1}{4\sqrt{2}\times 3.14{\left(1.2\times {10}^{-10}\right)}^2\times 1.2\times {10}^{26}}\\ \lambda \approx 3.3\times {10}^{-8}\text{m}\end{array}$

The average distance for oxygen is $5.6\times {10}^{-8}\text{m}$which is more than the average distance for nitrogen particles. The average distance each molecule of nitrogen travels in Titan's atmosphere is about $1.7$time less than the average distance for oxygen.

The average distance that a nitrogen molecule travels between collisions is not reasonable because it is more than the diameter of the sphere. It shows that there will be no collision between the particles in the atmosphere of Titan.

Conclusion:

Thus, the average distance that a nitrogen molecule travels between collisions on Titan is $3.3\times {10}^{-8}\text{m}$and it is about $1.7$times less than the average distance for oxygen. This result is unreasonable.