Question

Consider the following properties of the atmosphere of the planet Mars at a particular measurement point on the surface, as measured by the Mars Rover: Average surface pressure: 6.1 mbar Average temperature: 210 K Atmospheric gas composition (by volume): carbon dioxide (CO2) 95.32%; nitrogen (N2) 2.7%; argon (Ar) 1.6%; oxygen (O2) 0.13%; carbon monoxide (CO) 0.08%.
a. What is the partial pressure of CO2 in the Martian atmosphere, pA, in units of Pa?
b. What is the molar concentration of CO2 in the Martian atmosphere, cA?
c. What is the total molar concentration of all gases in the Martian atmosphere?
d. What is the total mass concentration of all gases the Martian atmosphere?

Answer 1

Answer:

a. 581.4 Pa

b. 3.33x10⁻⁴ mol/L

c. 3.49x10⁻⁴ mol/L

d. 0.015 g/L

Explanation:

a. By the Raoult's Law, the partial pressure of a component of a gas mixture is its composition multiplied by the total pressure, so:

pA = 0.9532*6.1

pA = 5.81452 mbar

pA = 5.814x10⁻³ bar

1 bar ----- 10000 Pa

5.814x10⁻³ bar--- pA

pA = 581.4 Pa

b. Considering the mixture as an ideal gas, let's assume the volume as 1,000 L, so by the ideal gas law, the total number of moles is:

PV = nRT

Where P is the pressure (610 Pa), V is the volume (1 m³), n is the number of moles, R is the gas constant (8.314 m³.Pa/mol.K), and T is the temperature.

n = PV/RT

n = (610*1)/(8.314*210)

n = 0.3494 mol

The number of moles of CO₂ is (V = 0.9532*1 = 0.9532 m³):

n = PV/RT

n = (581.4*0.9532)/(8.314*210)

n = 0.3174 mol

cA = n/V

cA = 0.3174/953.2

cA = 3.33x10⁻⁴ mol/L

c. c = ntotal/Vtotal

c = 0.3494/1000

c = 3.49x10⁻⁴ mol/L

d. The molar masses of the gases are:

CO₂: 44 g/mol

N₂: 28 g/mol

Ar: 40 g/mol

O₂: 32 g/mol

CO: 28 g/mol

The molar mass of the mixture is:

M = 0.9532*44 + 0.027*28 + 0.016*40 + 0.0008*28 = 43.36 g/mol

The mass concentration is the molar concentration multiplied by the molar mass:

3.49x10⁻⁴ mol/L * 43.36 g/mol

0.015 g/L