The usual units for molar mass M are grams per mole; for example, the molar mass

Question

The usual units for molar mass M are grams per mole; for example, the molar mass of oxygen is 32 { m g/mol}. These units are often omitted in tables. When you use SI units in equations, such as the equation v_{ m rms}=sqrt {{3 R T}/{M}}, you must express M in kilograms per mole by multiplying the value shown in the table by the ratio 1 ; m kg/10^3 ; m g. Thus, in SI units, M for oxygen is 32 imes 10^{-3} ; { m kg/mol}.
Determine whether you are working on a "per molecule" basis or a "per mole" basis. Keep in mind that m is the mass of a molecule and M is the mass of a mole; N is the number of molecules, and n is the number of moles; k is the gas constant per molecule, and R is the gas constant per mole.
Remember that T must always be in kelvins. If you are given temperatures in degrees Celsius, be sure to add 273 to convert to kelvins.

SET UP
Before writing any equations, organize your information and draw appropriate diagrams.
Part A
Start this problem by identifying the molar masses M_{ m N_2} and M_{ m H_2}, of nitrogen molecules ( m N_2) and hydrogen molecules ( m H_2) respectively, and write down their values.
Enter your answers in kilograms per mole to three significant figures, separated by a comma.
M_{ m N_2},M_{ m H_2} =

Part B
What is the most convenient approach to the problem? In the following expressions, refer to the notation used in the "problem-solving strategy."
A) A "per molecule" approach, where the equation v_{ m rms}=sqrt {rac {3 k T}{m}}is used to calculate the root-mean-square speed of a molecule.
B) A "per molecule" approach, where the equation v_{ m rms}=sqrt {rac {3 R T}{M}}is used to calculate the root-mean-square speed of a molecule.
C) A "per mole" approach, where the equation v_{ m rms}=sqrt {rac {3 k T}{m}}is used to calculate the root-mean-square speed of a molecule.
D) A "per mole" approach, where the equation v_{ m rms}=sqrt {rac {3 R T}{M}}is used to calculate the root-mean-square speed of a molecule.

Part C
What is the temperature in kelvins, T_H_2, of hydrogen molecules at 20.0 degrees C?
Enter your answer in kelvins to four significant figures.
T_H_2 =


Now that you have set up the problem, choose appropriate equations and solve for your unknowns.

Part D
At what temperature T_N_2 is the root-mean-square speed of nitrogen molecules equal to the root-mean-squar277e speed of hydrogen molecules at 20.0 degrees C?
Express your answer in degrees Celsius to four significant figures.
T_N_2 =

Part E
If nitrogen and hydrogen molecules are at the same temperature, how do their root-mean-square speeds compare?
A) v_rms of N_2 is lower than v_rms of H_2.
B) v_rms of N_2 is equal to v_rms of H_2.
C) v_rms of N_2 is higher than v_rms of H_2.

Explanation / Answer

MOLAR MASSES IN KILOGRAMES PER MOLES

Molar mass of nitrogen N2 = 28 g / mol

Molar mass of Hydrogen H2 = 2.015g/mol

Molar mass of oxygen O2 = 32g / mol

MOLAR MASSES IN Kg / mol

Molar mass of hydrogen H2 = 2.015(10-3) Kg / mol = 0.002015 Kg / mol

N2 = 28(10-3) Kg / mol = 0.028 Kg / mol

O2 = 32 (10-3) kg / mol = 0.032 Kg / mol

PART B

The approch must be

option A .   Becasue K and m are used in terms of molecules

Option D Because R and M are used in terms of moles.

Part C

T in kelvin = ?

T = 20 +273 = 293 K

Part E

Root mean square formula for velocity is

Vrms of H2 = (3RTh / Mh)1/2

Vrms of N2 = (3RTn / Mn)1/2

As it is givne Vrms of H2 = Vrms of N2

(3RTh / Mh)1/2 =( 3RTn / Mn)1/2

Th / Mh = Tn / Mn

Tn = Th Mn / Mh

Tn = 20 ( 0.028 /0.002015)

Tn = 277.91 C

Part E

IF they both are at same temprature then the vrms of nitrogen will be less than that of hydrogen because nitrogen molecuales are heavier than hydrogen molecules. So, option A is correct one.

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