Question

Space probes to Mars have shown that its atmosphere consists mostly of carbon dioxide. The average temperature on the surface of Mars is \(-55^{\circ} \mathrm{C}\) with an average pressure of 0.00592 atm. Compare the density of \(\mathrm{CO}_{2}\) on Mars's surface with that on the earth's surface at \(25^{\circ} \mathrm{C}\) and one atmosphere.

Short answer

Answer: The density of CO2 on Mars's surface is approximately 8.86 times greater than that on Earth's surface at 25°C and one atmosphere.

Step-by-step solution

Identify the temperature and pressure values for both cases

For Mars, the temperature T_Mars is \(-55^{\circ}\mathrm{C}\), and the pressure P_Mars is 0.00592 atm. To work with these values, we need to convert them to Kelvins and Pascals, respectively. For Earth, the temperature T_Earth is \(25^{\circ}\mathrm{C}\), and the pressure P_Earth is 1 atm.

Convert Celsius to Kelvin

Convert the temperatures from Celsius to Kelvin using the formula K = °C + 273.15. T_Mars = \( (-55 + 273.15) \mathrm{K} = 218.15 \mathrm{K}\) T_Earth = \((25 + 273.15) \mathrm{K} = 298.15 \mathrm{K}\)

Convert atm to Pascal

Convert the pressures from atm to Pascal (Pa) using the conversion factor 1 atm = 101325 Pa. P_Mars = \((0.00592 \times 101325) \mathrm{Pa} = 599.84 \mathrm{Pa}\) P_Earth = \((1 \times 101325) \mathrm{Pa} = 101325 \mathrm{Pa}\)

Calculate the molar mass of CO2

The molar mass of \(\mathrm{CO}_{2}\) (M) is the sum of the molar masses of carbon (C) and two oxygen (O) atoms: M = \(12.01 \ \mathrm{g/mol (C)} + 2 \times 16.00 \ \mathrm{g/mol (O)} = 44.01 \ \mathrm{g/mol}\)

Use the ideal gas law to calculate the density

The ideal gas law is given by \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Rearranging and solving for density (ρ), we get: \(\rho = \frac{nM}{V} = \frac{PM}{RT}\) For Mars: \(\rho_\mathrm{Mars} = \frac{599.84 \ \mathrm{Pa} \times 44.01 \ \mathrm{g/mol}}{(8.314 \ \mathrm{J/(mol \cdot K)} \times 218.15 \ \mathrm{K}} = 16.476 \ \mathrm{kg/m^3}\) For Earth: \(\rho_\mathrm{Earth} = \frac{101325 \ \mathrm{Pa} \times 44.01 \ \mathrm{g/mol}}{(8.314 \ \mathrm{J/(mol \cdot K)} \times 298.15 \ \mathrm{K}} = 1.859 \ \mathrm{kg/m^3}\)

Compare the densities

To compare the densities, calculate the ratio of the density on Mars to the density on Earth: Ratio = \(\frac{\rho_\mathrm{Mars}}{\rho_\mathrm{Earth}} = \frac{16.476 \ \mathrm{kg/m^3}}{1.859 \ \mathrm{kg/m^3}} = 8.86\) The density of \(\mathrm{CO}_{2}\) on Mars's surface is approximately 8.86 times greater than that on Earth's surface at \(25^{\circ}\mathrm{C}\) and one atmosphere.

Key concepts

Ideal Gas Law

Understanding the ideal gas law is crucial when comparing atmospheric densities on different planets. This fundamental equation relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas. Expressed as \(PV = nRT\), where R is the universal gas constant, it provides insights into how gas behaves under various conditions of temperature and pressure.

When applying the ideal gas law, a common task is to find the density of the gas. By manipulating the equation, we can solve for density (\(\rho\)), represented by the mass (m) of gas per unit volume (V). Since density is also the product of molar mass (M) and number of moles (n) divided by volume, we can write \(\rho = \frac{nM}{V}\). Substituting the ideal gas law gives us the formula \(\rho = \frac{PM}{RT}\), which was used in the exercise to compare the densities of Mars and Earth's atmospheres.

Conversion of Temperature Units

Temperature conversions are an important step in gas calculations, particularly when using the ideal gas law. Since this law requires temperature to be in Kelvin, we must convert Celsius to Kelvin for accurate calculations. The conversion is straightforward: \(K = \text{°C} + 273.15\).

This conversion crucially ensures that temperature is measured on an absolute scale, where 0 K (or absolute zero) is the theoretical point where particles have minimal thermal motion. By converting temperatures to Kelvins, we maintain consistency with scientific standards and achieve reliable results in our gas law calculations.

Conversion of Pressure Units

Pressure conversion is another key aspect of gas calculations since the ideal gas law requires pressure to be in consistent units. In the context of our exercise, we need to convert atmospheric pressure (atm) to Pascals (Pa). The conversion factor is 1 atm equals 101325 Pa.

It is essential to use the correct unit for pressure when applying the ideal gas law to ensure the units of the gas constant and other terms in the equation match up. This consistency across units allows precise comparisons and calculations of gas properties.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of its particles and is central to calculations involving gas density and molar quantities. To calculate the molar mass of a compound, one sums the molar masses of its constituent elements, each multiplied by the number of atoms of that element in one formula unit of the compound.

In our exercise, we calculate the molar mass of carbon dioxide (\(\mathrm{CO}_{2}\)) by adding the molar mass of one carbon atom (12.01 g/mol) to the molar mass of two oxygen atoms (2 x 16.00 g/mol), yielding a molar mass of 44.01 g/mol for \(\mathrm{CO}_{2}\).

Gas Density Calculation

Gas density is the measure of a gas's mass per unit volume and is vital for comparing planetary atmospheres. It's obtained by combining the molar mass of the gas with the ideal gas law rearranged to express density, \(\rho = \frac{PM}{RT}\), where \(P\) is the pressure, \(M\) is the molar mass, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. This formula allows us to calculate the density directly if the other quantities are known.

In our exercise comparing Mars's atmosphere with Earth's, understanding gas density helps highlight the stark differences between both planets. Despite Mars having a much thinner atmosphere, its \(\mathrm{CO}_{2}\) density is significantly higher at the surface due to the lower temperature and the fact that its atmosphere is predominantly \(\mathrm{CO}_{2}\).