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Tems11 [23]
2 years ago
11

Consider 3.01*10^24 molecules of oxygen (atomic mass is 16.0 g/mole) in a container at a temperature of 22.0 degrees celsius. If

this gas can be considered as an ideal gas, calculate: 1.The Average translational kinetic energy of the gas. 2.The average translational kinetic energy per molecule 3. The root mean square speed of a gas molecule. 4. The most probable speed of the gas molecules.
Chemistry
1 answer:
Sergeeva-Olga [200]2 years ago
8 0

Answer:

See explanation

Explanation:

The tranlational energy of a molecule is obtained by; ET =3/2 kT

T is the temperature while k is the Boltzmann constant (k)

The Boltzmann constant (kB) relates temperature to energy.

Substituting values;

ET = 3/2 (1.3807 × 10−23 joule/K) * (22 + 273)K

ET = 6.11 * 10^-21 J

The average translational kinetic energy per molecule=  6.11 * 10^-21 J/ 3.01*10^24 molecules = 2.03 * 10^-45 J per molecule

<u>vrms = √</u>3RT/M

vrms = √3 * 8.314 * 295/32

vrms = 15.2 m/s

Most probable speed =  √2RT/M

Most probable speed = √2 * 8.314 * 295/32

Most probable speed = 12.38 m/s

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A certain liquid X has a normal boiling point of 111.20 celsius and a boiling point elevation constant Kb=0.95. calculate the bo
Goshia [24]

Answer: the boiling point is = 137.325°C

Explanation:

From the formula: ∆Tb= Kb*m

From the question, Kb= 0.95, m= 27.5, T1= 111.2°C

Substitute into ∆Tb= Kb*m

∆Tb= 0.95*27.5= 26.125

∆Tb= T2-T1

Hence

T2- 111.2=26.125

T2= 26.125+ 111.2= 137.325°C

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Explanation:

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2 years ago
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Shtirlitz [24]

Answer:

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Explanation:

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