This is a physical Chemistry question, I need help with all parts. The parts fro

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This is a physical Chemistry question, I need help with all parts. The parts from equation 17.25 are mass of the earth, gravitational constant (G), and the radius of the earth.

5. Let's apply kinetic theory to a real-world problem: Why does the Earth's atmosphere have the composition that it does? a) Calculate the most probable speed, average speed, and root-mean-square speed for Hew at 25° C. Report your answers in km/s b) The minimum speed necessary for an object to leave the influence of Earth's gravity and never return is known as the eseape velocity. Use the information in Problem 17.25 to calculate the escape velocity for the Earth in km/s. Note that the escape velocity is independent of the mass of the object that escapes. (c) A planet can hold a gas in its atmosphere if its escape velocity is at least 10 times higher than the average speed of molecules in the gas at the appropriate temperature. Otherwise, the gas will leak into space. Use this information to explain why He (and similarly, H2) is not found in significant quantities in Earth's atmosphere. Assume that the temperature of the atmosphere is 25o C d) Calculate the molecular weight of the lightest gas that could theoretically be a significant part of Earth's atmosphere. Ignore the question of whether or not a gas with this molecular weight actually exists

Explanation / Answer

a) rms speed = sqrt (3RT/M)

Where R = 8.314 J/mol.K = 8.314 N.m/mol.K = 8.314 Kg.m2/s2.mol.K

T = 298 K

Molar mass of He = 4.0 x 10^-3 Kg/mol

rms speed for He = sqrt( 3RT / M )

sqrt( (3 x 8.314 x 295) / 4.00 e-3 ) = 1356.24 m/s = 1.36 Km/s

Most probable speed for He = sqrt (2RT/M)

= sqrt(2 x 8.314 x 295) / 4.00 e-3 ) = 1107.30 m/s = 1.1 Km/s

Average speed for He = sqrt (8RT/M)

sqrt( (8 x 8.314 x 295) / 3.14 x 4.00 e-3 ) = 1250 m/s = 1.25 Km/s

b) The equation for

Escape velocity = Sqrt(2GM/r)

G is gravitational constant has the value: G = 6.67 * 10-11 Newton-meter2/kilogram2.

M is mass of earth = 5.98 × 10^24 kg

r is radius of earth = 6.38 x 10^6 m

Substitute all the value in equation

Escape velocity = Sqrt [(2(6.67 * 10^-11 m3/Kgs2)( 5.98 × 10^24 kg)/ (6.38 x 10^6 m)]

             = Sqrt [(2(6.67 * 10^-11)( 5.98 × 10^24 k)/ (6.38 x 10^6 )m2/s2]

              = 11200 m/s

              = 11.2 Km/s

c) Ratio between Escape velocity/ Average speed for He = 11.2 Km/s/1.25 Km/s = 8.96

A planet can hold a gas in its atmosphere if its escape velocity is at least 10 times higher than the average speed of molecules in the gas at the appropriate temperature. Otherwise, the gas will leak into space. In case of Helium escape velocity is only 8.96 times higher than the average speed of Helium at the appropriate temperature. So Helium can escape easily from earth’s atmosphere and not found in significant quantities in Earth’s atmosphere.

d) Mass of lightest gas that is part of earth’s atmosphere

Average speed for lightest = 11.2/10 = 1.12 Km/s = 1120 m/s

Average speed for = sqrt (8RT/M)

sqrt (M) = Average speed /sqrt(8RT/)

sqrt (M) = (1120 )/sqrt( (8 x 8.314 x 295) / 3.14 )

sqrt (M) = sqrt (M) = sqrt (31.62)

M = 5.6 g/mol

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