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

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

i believe the answer is three water molecules

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.