Chapter 16, Problem 50P

(a)

To determine

Theoretical lapse rate on the Earth.

Answer to Problem 50P

Theoretical lapse rate on the Earth is $\underset{\_}{-9.73°\text{C}/\text{km}}$.

Explanation of Solution

Given that, the expression for lapse rate.

    $\frac{dT}{dy}=\frac{\gamma -1}{\gamma }\frac{gM}{R}$        (I)

Here, $\frac{dT}{dy}$ is the lapse rate, $\gamma$ is the ratio of specific heat, $g$ is the acceleration due to gravity, $R$ is the universal gas constant, and $M$ is the molar mass of the uniform ideal gas.

Conclusion:

Substitute, $1.40$ for $\gamma$, $9.80\text{m}/{\text{s}}^2$ for $g$, $28.9\text{g}/\text{mol}$ for $M$, and $8.314\text{J}/\text{mol}\cdot \text{K}$ in the equation (I), to find $\frac{dT}{dy}$.

    $\begin{array}{c}\frac{dT}{dy}=\frac{1.40-1}{1.40}\frac{\left(9.80\text{m}/{\text{s}}^2\right)\left(28.9\text{g}/\text{mol}\right)}{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ =\frac{0.40}{1.40}\frac{\left(9.80\text{m}/{\text{s}}^2\right)\left(28.9\text{g}/\text{mol}\right)}{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\left(\frac{1\text{kg}}{1000\text{g}}\right)\left(\frac{1\text{J}}{1\text{kg}\cdot {\text{m}}^2/{\text{s}}^2}\right)\\ =-9.73\times {10}^{-3}\text{K}/\text{m}\\ =-9.73°\text{C}/\text{km}\end{array}$

Therefore, theoretical lapse rate on the Earth is $\underset{\_}{-9.73°\text{C}/\text{km}}$.

(b)

To determine

Reason for the theoretical lapse rate is different from the value $-6.5°\text{C}/\text{km}$.

Answer to Problem 50P

Air contains water vapor. This water vapor condense out as liquid drops if the parcel of air rises in the atmosphere. So the heat of vaporization due to this condensation will result in the raising temperature than the value $-6.5°\text{C}/\text{km}$.

Explanation of Solution

The air does not behaves as an ideal gas since it contain water vapor. When the parcel of air roses in the atmosphere, its temperature decreases. Then the ability to contain water vapor decreases. So the water vapor condense out as liquid drops. It will produce heat if vaporization. Hence the temperature of the air increases above the value $-6.5°\text{C}/\text{km}$.

Conclusion:

Therefore, the theoretical lapse rate is different from the value $-6.5°\text{C}/\text{km}$.

(c)

To determine

Lapse rate for the Martian troposphere.

Answer to Problem 50P

Lapse rate for the Martian troposphere is $\underset{\_}{-4.60°\text{C}/\text{km}}$.

Explanation of Solution

Write the expression for acceleration due to gravity in the Mars.

    $g=\frac{G{M}_{\text{Mars}}}{r_{\text{Mars}}^2}$        (II)

Here, $g$ is the acceleration due to gravity, $G$ is the gravitational constant, $M_{\text{Mars}}$ is the mass of the Mars, and $r_{\text{Mars}}$ is the radius of the Mars.

Conclusion:

Substitute, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $6.42\times {10}^{23}\text{kg}$ for $M_{\text{Mars}}$, and $3.37\times {10}^6\text{m}$ for $r$ in the equation (II), to find $g$.

    $\begin{array}{c}g=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(6.42\times {10}^{23}\text{kg}\right)}{{\left(3.37\times {10}^6\text{m}\right)}^2}\\ =3.77\text{m}/{\text{s}}^2\end{array}$

Substitute, $1.30$ for $\gamma$, $3.77\text{m}/{\text{s}}^2$ for $g$, $0.0440\text{kg}/\text{mol}$ for $M$, and $8.314\text{J}/\text{mol}\cdot \text{K}$ in the equation (I), to find $\frac{dT}{dy}$.

    $\begin{array}{c}\frac{dT}{dy}=\frac{1.30-1}{1.30}\frac{\left(3.77\text{m}/{\text{s}}^2\right)\left(0.0440\text{kg}/\text{mol}\right)}{\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ =-4.60\times {10}^{-3}\text{K}/\text{m}\\ =-4.60°\text{C}/\text{km}\end{array}$

Therefore, lapse rate for the Martian troposphere is $\underset{\_}{-4.60°\text{C}/\text{km}}$.

(d)

To determine

The height in the Martial troposphere at which the temperature is $-60.0°\text{C}$.

Answer to Problem 50P

The height in the Martial troposphere at $-60.0°\text{C}$ is $\underset{\_}{4.34\text{km}}$.

Explanation of Solution

The lapse rate can be expressed as,

    $\frac{\Delta T}{\Delta y}=\frac{dT}{dy}$        (III)

Solve the equation (III) for $\Delta y$.

    $\Delta y=\frac{\Delta T}{\frac{dT}{dy}}$        (IV)

Conclusion:

Substitute, $-60.0°\text{C}-\left(-40.0°\text{C}\right)$ for $\Delta T$, and $-4.60°\text{C}/\text{km}$ for $\frac{dT}{dy}$ in the equation (IV), to find $\frac{dT}{dy}$.

    $\begin{array}{c}\Delta y=\frac{-60.0°\text{C}-\left(-40.0°\text{C}\right)}{-4.60°\text{C}/\text{km}}\\ =4.34\text{km}\end{array}$

Therefore, the height in the Martial troposphere at $-60.0°\text{C}$ is $\underset{\_}{4.34\text{km}}$.

(e)

To determine

The reason for the dust affect the lapse rate, and the mission occurred in dustier conditions, Mariner or viking.

Answer to Problem 50P

The dust in the atmosphere contribute energy to the gas molecule, it will results the increases in internal energy of the atmosphere, and decrease in temperature with height. If the dust rate is larger, then the lapse rate will deviate from the theoretical value. This was the dustier during the Marine fights.

Explanation of Solution

The dust in the atmosphere absorbs and scatters energy from the electromagnetic radiation coming from the sun. The internal energy of the atmosphere increases, and temperature of the air decreases with height due to this contribution of the dust energy to the gas. If the dust does not absorb sunlight, the rate of decrease of temperature is less than previous case.so atmosphere contain more dust lapse rate will be more, and it will deviate from the theoretical value. This was dustier during the Mariner flights.

Conclusion:

Therefore, lapse rate will deviate from the theoretical value, and this was the dustier during the Marine fights.