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Helium (He) is the lightest noble gas component of air, and xenon (Xe) is the heaviest. Perform the following calculations, using R = 8.314 J/(mol·K) and ℳ in kg/mol. (a) Find the rms speed of He in winter (0.°C) and in summer (30.°C). Enter your answers in scientific notation. × 10 m/s (winter) × 10 m/s (summer) (b) Find the ratio of the rms speed of He to that of Xe at 30.°C. (rate He)/(rate Xe) (c) Find the average kinetic energy per mole of He and of Xe at 30.°C. Enter your answers in scientific notation. × 10 J/mol for He × 10 J/mol for Xe (d) Find the average kinetic energy per molecule of He at 30.°C. Enter your answer in scientific notation. × 10 J/He atom

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Answer:

Explanation:

Hello,

Data;

R = 8.314J/(mol.K)

Temp (winter) = 0°C = (0 + 273.15)K = 274.15K

Temp.(summer) = 30°C = (30 + 273.15)K = 303.15K

Molar mass of He = 4g/mol = 0.004kg/mol

Molar mass of Xe = 131.29g/mol = 0.131kg/mol

a) rms speed in winter and summer

Vrms = √(3RT/M)

R = gas constant

T = temperature of the gas

M = molar mass of the gas

In winter,

Vrms = √(3×8.314×273.15) / 0.004

Vrms = 1.30×10³m/s

In summer

Vrms = √(3×8.314×303.15) / 0.004

Vrms = 1.37×10³m/s

b) Vrms of Xe at 30°C

Vrms = √(3 × 8.314 × 303.15) / 0.131

Vrms of Xe = 240.24m/s = 2.40×10²m/s

Vrms of He = 1.37×10³m/s

Rate of He / rate of Xe = 1.37×10³ / 2.40×10²

Rate of He / rate of Xe = 5.7

c) K.E per mole

At 30°C

K.E of He = (3/2) × 8.314 × 303.15

K.E of He = 3.78×10³J/mol

K.E of Xe = (3/2) × 8.314 × 303.15

K.E of Xe = 3.78×10³J/mol

d) K.E per molecule = ½mv²

K.E of He molecule = ½ × 0.004 × 1.37×10³

K.E = 2.74J

The rms speed of He in winter and in summer are 1.30×10³m/s & 1.37×10³m/s respectively, ratio of the rms speed of He to that of Xe at 30 degree celsius is 5.7, average kinetic energy per mole of He and of Xe and average kinetic energy per molecule of He are discussed below.

How do we calculate the root mean square velocity?

Root mean square velocity of the gases will be calculated as:

Vrms = √(3RT/M), where

R = universal gas constant = 8.314J/(mol.K)

M = molar mass of gas in kg/mol

T = temperature in K

  • Root mean square velocity of Helium at 0°C or 273.15K and 30°C or 303.15K will be calculated as:

At 273.15K-

Vrms = √(3×8.314×273.15) / 0.004 = 1.30×10³m/s

At 303.15K-

Vrms = √(3×8.314×303.15) / 0.004 = 1.37×10³m/s

Vrms (winter×summer) = 1.30×10³m/s × 1.37×10³m/s = 1.78×10⁶m/s

  • Vrms of Xe at 303.15K

Vrms = √(3 × 8.314 × 303.15) / 0.131

Vrms of Xe = 240.24m/s = 2.40×10²m/s

Vrms of He = 1.37×10³m/s

Rate of He / rate of Xe = 1.37×10³ / 2.40×10²

Rate of He / rate of Xe = 5.7

  • Average kinetic energy will be calculated as:

K.E = 3RT / 2N, where

N = avgadros number = 6.022×10²³ atoms/mole

Average kinetic energy per mole of He & Xe at 303.15K is same as:

K.E = 3/2(1.38 × 10⁻²³J/K)(303.15K) = 6.2 × 10⁻²⁵J/mol

  • Average kinetic energy per molecule of He will be calculated as:

K.E of He molecule = ½ × 0.004 × 1.37×10³

K.E = 2.74J

Hence, calculations for the given points are described above.

To know more about root mean square velocity, visit the below link:

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