1.00 mole of gaseous water (H2O) and 1.00 mole of gaseous carbon dioxide (CO2) both have the same kinetic energy at 100.0ºC because gas kinetic energies depend on absolute temperature.

Question

1.00 mole of gaseous water (H2O) and 1.00 mole of gaseous carbon dioxide (CO2) both have the same kinetic energy at 100.0ºC because gas kinetic energies depend on absolute temperature.

How do their average molecular velocities compare, given urms = √(3RT/Mwt)?

Mwts: H = 1.0 g; C = 12.0 g; O = 16.0 g

Question 1Select one:

a.
Their average molecular velocities are also equal.

b.
The water average molecular velocity is larger than the carbon dioxide average molecular velocity.

c.
The carbon dioxide average molecular velocity is larger than the water average molecular velocity.

d.
There is not enough information to answer this question.

Answer 1

To compare the average molecular velocities, we can use the equation:

urms = √(3RT/Mwt)

The formula shows that the average molecular velocity is inversely proportional to the square root of the molecular weight (Mwt).

Since carbon dioxide (CO2) has a higher molecular weight (44 g/mol) compared to water (H2O) (18 g/mol), the water average molecular velocity will be larger than the carbon dioxide average molecular velocity.

Therefore, the correct answer is: b. The water average molecular velocity is larger than the carbon dioxide average molecular velocity.

Answer 2

To compare the average molecular velocities of gaseous water and gaseous carbon dioxide, we can use the equation:

urms = √(3RT/Mwt)

where:
urms is the root mean square velocity
R is the gas constant (8.314 J/(mol·K))
T is the absolute temperature in Kelvin (which can be calculated as 100.0ºC + 273.15 = 373.15 K)
Mwt is the molar mass of the gas in grams/mol

Let's calculate the average molecular velocities for water and carbon dioxide:

Water (H2O):
Mwt = 2(1.0 g/mol of hydrogen) + 16.0 g/mol of oxygen = 18.0 g/mol

urms_water = √(3(8.314 J/(mol·K))(373.15 K)/(18.0 g/mol))
urms_water ≈ 1,880 m/s

Carbon dioxide (CO2):
Mwt = 12.0 g/mol of carbon + 2(16.0 g/mol of oxygen) = 44.0 g/mol

urms_carbon_dioxide = √(3(8.314 J/(mol·K))(373.15 K)/(44.0 g/mol))
urms_carbon_dioxide ≈ 513 m/s

Comparing the values, we find that the water average molecular velocity (urms_water) is significantly larger than the carbon dioxide average molecular velocity (urms_carbon_dioxide). Therefore, the correct answer is:

b. The water average molecular velocity is larger than the carbon dioxide average molecular velocity.

1.00 mole of gaseous water (H2​O) and 1.00 mole of gaseous carbon dioxide (CO2) both have the same kinetic energy at 100.0ºC because gas kinetic energies depend on absolute temperature.

To compare the average molecular velocities, we can use the equation:

To compare the average molecular velocities of gaseous water and gaseous carbon dioxide, we can use the equation:

1.00 mole of gaseous water (H2O) and 1.00 mole of gaseous carbon dioxide (CO2) both have the same kinetic energy at 100.0ºC because gas kinetic energies depend on absolute temperature.