Question

The atmosphere of Mars exerts a pressure of only 600 . Pa on the surface and has a density of only \(0.0200 \mathrm{~kg} / \mathrm{m}^{3}\) a) What is the thickness of the Martian atmosphere, assuming the boundary between atmosphere and outer space to be the point where atmospheric pressure drops to \(0.0100 \%\) of its value at surface level? b) What is the atmospheric pressure at the bottom of Mars's Hellas Planitia canyon, at a depth of \(8.18 \mathrm{~km} ?\) c) What is the atmospheric pressure at the top of Mars's Olympus Mons volcano, at a height of \(21.3 \mathrm{~km} ?\) d) Compare the relative change in air pressure, \(\Delta p / p\), between these two points on Mars and between the equivalent extremes on Earth - the Dead Sea shore, at 400 , \(\mathrm{m}\) below sea level, and Mount Everest, at an altitude of \(8850 \mathrm{~m}\).

Short answer

Answer: The approximate ratio of the relative change in air pressure on Mars compared to Earth is 3.47.

Step-by-step solution

Find the mass of the Martian atmosphere

We can find the mass of the Martian atmosphere using the given density and pressure: $$ m = \frac{P}{g \rho} $$ Where \(P = 600\,\text{Pa}\), \(g = 3.711\,\text{m/s}^2\) (gravity on Mars), and \(\rho = 0.0200\,\text{kg/m}^3\) $$ m = \frac{600}{(3.711)(0.0200)} = 8074.89\,\text{kg/m}^2 $$

Apply the ideal gas law

We can now use the ideal gas law to find the volume of the Martian atmosphere: $$ PV=nRT \\ P = \frac{nRT}{V} $$ Where \(R = 8.314\,\text{J/(mol K)}\) is the gas constant, \(T = 210\,\text{K}\) is the temperature of Mars' atmosphere (approximate), and \(n =\) moles of gas. We need to convert the mass to moles, so we'll use the molar mass of Mars' atmosphere, which is primarily CO2 with a molar mass of \(44.01\,\text{g/mol}\) or \(0.04401\,\text{kg/mol}\): $$ n = \frac{8074.89}{0.04401} = 183477.85\,\text{mol} $$ Now, we can solve for the volume (V): $$ V=\frac{nRT}{P} \\ V=\frac{(183477.85)(8.314)(210)}{600}= 5.363\times10^{6}\,\text{m}^3 $$

Solve for the height of the atmosphere

We can now solve for the height (thickness) of the Martian atmosphere, assuming it has a uniform density and a base area (A) equal to the surface area of Mars. $$ V = A \times h \\ h = \frac{V}{A} $$ The surface area of Mars is \(A = 1.452\times10^{14}\,\text{m}^2\). So the thickness (h) of the atmosphere is: $$ h = \frac{5.363\times10^{6}}{1.452\times10^{14}} = 3.69\times10^{-8}\,\text{m} $$ b) Find the atmospheric pressure at the bottom of Mars's Hellas Planitia canyon:

Find the additional pressure

Calculate the additional pressure due to the depth of the canyon using the atmospheric density and gravity on Mars: $$ \Delta P = \rho g h \\ \Delta P = (0.0200)(3.711)(8180) = 607.554\,\text{Pa} $$

Calculate the total pressure

Add the additional pressure to the atmospheric pressure to find the total pressure at the bottom of Hellas Planitia canyon: $$ P_{total} = P_{surface} + \Delta P \\ P_{total} = 600 + 607.554 = 1207.554\,\text{Pa} $$ c) Find the atmospheric pressure at the top of Mars's Olympus Mons volcano:

Find the additional pressure

Calculate the additional pressure due to the height of the volcano: $$ \Delta P = \rho g h \\ \Delta P = (0.0200)(3.711)(21300) = 1578.774\,\text{Pa} $$

Calculate the total pressure

Subtract the additional pressure from the atmospheric pressure to find the total pressure at the top of Olympus Mons: $$ P_{total} = P_{surface} - \Delta P \\ P_{total} = 600 - 1578.774 = -978.774\,\text{Pa} $$ Since the pressure cannot be negative, we conclude that the pressure at the top of Olympus Mons is approximately zero. d) Compare the relative change in air pressure on Mars and Earth:

Calculate the change in pressure on Mars

Calculate the difference in pressure between Hellas Planitia and Olympus Mons and divide by the initial pressure (at surface level): $$ \frac{\Delta P}{P} = \frac{1207.554}{600} = 2.0126 $$

Calculate the change in pressure on Earth

Perform a similar calculation for Earth's locations, using the given altitudes and Earth's standard atmospheric pressure (\(101325\,\text{Pa}\)): $$ \Delta P_{\text{Dead Sea}} = 400\,\text{m} \times 1.225\,\text{kg/m}^3 \times 9.81\,\text{m/s}^2 = 4808.7\,\text{Pa} \\ \Delta P_{\text{Everest}} = 8850\,\text{m} \times 1.225\,\text{kg/m}^3 \times 9.81\,\text{m/s}^2 = 106971.175\,\text{Pa} \\ $$ $$ \frac{\Delta P}{P} = \frac{106971.175 - 4808.7}{101325} = 0.5805 $$

Compare the results

Compare the relative change in air pressure between Mars and Earth: The relative change in air pressure on Mars is 2.0126, whereas on Earth it is 0.5805. This means that the relative change in air pressure on Mars is approximately 3.47 times greater than the relative change in air pressure on Earth.

Key concepts

Martian Atmospheric Composition

Understanding the atmospheric composition of Mars is crucial for several scientific and practical reasons, including the study of its climate, the potential for future human exploration, and the challenges it poses for spacecraft design. The Martian atmosphere is vastly different from Earth's, with carbon dioxide (CO₂) comprising about 95.3% of the gas present. The remainder includes nitrogen (2.7%), argon (1.6%), and traces of oxygen and water vapor.

The predominance of CO₂ places importance on understanding the properties of this gas under Martian conditions, especially for the ideal gas law application. When solving exercises related to the Martian atmosphere, considering the specific gases present ensures a more accurate calculation of pressure, volume, and temperature conditions on the planet. Moreover, the low density of the atmosphere, approximately 100 times less dense than Earth's, significantly affects various physical processes and has implications for aerodynamic flight and heat transfer in the Martian environment.

Ideal Gas Law Application

The ideal gas law is a fundamental equation in thermodynamics, relating the pressure, volume, temperature, and amount of a gas. On Mars, applying the ideal gas law enables us to describe the behavior of the atmospheric gases under varying conditions. It is expressed as
\[PV = nRT\],
where P is the pressure, V is the volume, n represents the number of moles, R is the universal gas constant, and T is the temperature expressed in Kelvin.

For example, solving the exercises related to Martian atmosphere pressure, one must adapt the Earth-based constants in the ideal gas law to Martian specifics, such as its gravity and atmospheric composition. The use of an appropriate molar mass for the dominant Martian atmospheric gas, CO₂, is an essential aspect of obtaining an accurate solution. The law applies perfectly in theoretical scenarios; however, Mars' thin atmosphere introduces complexities, such as non-ideal behavior at low pressures that must be considered for precise scientific work.

Mars Surface Gravity Effects

Mars' gravity, at only 0.38 of Earth's, creates unique conditions for its atmosphere. This reduced gravity affects the way gases in the atmosphere exert pressure and how they are distributed across different altitudes. For instance, the atmospheric pressure on Mars at the surface is only about 600 Pascals, and gravity influences how this pressure changes with altitude.

In solving problems like the change of atmospheric pressure at different locations on Mars, we must account for surface gravity in the calculation. For example, in the Hellas Planitia canyon and Olympus Mons scenarios, the varying atmospheric pressures are computed considering the product of atmospheric density, gravity, and change in elevation (either depth or height). Understanding this concept is fundamental for interpreting atmospheric phenomena on Mars and for the design of missions and technology to navigate and explore the Martian surface. Gravity's effect on the distribution and pressure of the Martian atmosphere cannot be understated, even when considering theoretical exercises regarding interplanetary atmospheres.

Gravitational Influence on Atmospheric Dynamics

The interplay between Mars' lower gravity and its atmosphere leads to a larger scale height (the height over which the atmospheric pressure decreases by a factor of e), meaning that the Martian atmosphere is more extended vertically compared to Earth's. This expansion impacts everything from meteorological processes to the potential habitability of the planet.