The atmosphere on Mars is almost all carbon dioxide (CO2), and has a pressure of about 6.9×102 Pa, much less than that of Earth. At −22°C (a warm day on Mars), what is the mass of one cubic meter of the atmosphere? (Assume the atmosphere is completely CO2.)

I got 0.020 using the equation (1/2) m v^2 = (3/2) K T.
Am I right? If not, what do I need to do and what answer should I get?

To find the mass of one cubic meter of the atmosphere on Mars, you need to use the ideal gas law and the molar mass of carbon dioxide (CO2). The equation you mentioned, (1/2) m v^2 = (3/2) K T, is actually the kinetic energy equation for a single gas molecule, and it won't directly give you the mass of the entire atmosphere.

To calculate the mass of one cubic meter of the atmosphere, we need to use the ideal gas law, which states:

PV = nRT

Where:
P is the pressure of the gas (6.9×10^2 Pa)
V is the volume of the gas (1 m^3)
n is the number of moles of the gas
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature of the gas (-22°C = 251.15 K)

We can rearrange the equation to solve for n:

n = PV / RT

Now, let's calculate the number of moles (n) of CO2 in one cubic meter of the atmosphere on Mars:

n = (6.9×10^2 Pa) * (1 m^3) / ((8.314 J/(mol·K)) * (251.15 K))

n ≈ 0.03393 moles (rounded to 5 decimal places)

Next, we need to calculate the mass (m) of CO2 in one mole. The molar mass of CO2 can be found by adding the atomic masses of carbon (C) and two oxygen (O) atoms:

Molar mass of C = 12.01 g/mol
Molar mass of O = 15.999 g/mol

Molar mass of CO2 = (12.01 g/mol) + 2 * (15.999 g/mol)
Molar mass of CO2 ≈ 44.01 g/mol (rounded to 2 decimal places)

Finally, we can calculate the mass of one cubic meter of the atmosphere on Mars:

Mass = n * Molar mass

Mass = (0.03393 moles) * (44.01 g/mol)
Mass ≈ 1.49 g (rounded to 2 decimal places)

Therefore, the mass of one cubic meter of the atmosphere on Mars is approximately 1.49 grams.

In conclusion, using the ideal gas law and the molar mass, you can determine the mass of one cubic meter of the atmosphere on Mars. The equation (1/2) m v^2 = (3/2) K T is not directly applicable to this specific calculation.