Question

The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would oxygen molecules (molecular mass is equal to 32.0 g/mol) have an average velocity vrms equal to Earth’s escape velocity of 11.1 km/s

Answer 1

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)$v_{es} =v_{rms}$= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

$v_{rms} = \sqrt{ \frac{3RT}{M}}$

So, $v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}$

multiply each side by M_{o2}, so we have

$v_{rms,o2}^2 *M_{o2} =3RT_{o2}$

solving for temperature T_{o2}

$T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}$

In the question given,$v_{rms} =v_{es}$

$T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}$

T_{o2} =1.58*10^5 K

Answer 2

(a) The speed of the object in Km/h is 40,320 km/h.

(b) The temperature of oxygen molecule at the given average velocity is 160,936.6 K.

Speed of the object in km/h

The speed of the object in Km/h is calculated as follows;

v = 11.2 km/s x 3600 s/h

v = 40,320 km/h

Temperature of the oxygen molecule

The temperature of the oxygen molecule is calculated as follows;

$V_{rms} = \sqrt{\frac{3RT}{M} } \\\\v_{rms}^2 = \frac{3RT}{M} \\\\T = \frac{v_{rms}^2M}{3R}$

where;

• R is gas constant = 8.314 J/k.mol
• M is molar mass of oxygen gas = 32 g/mol = 0.032 kg/mol

$T = \frac{(11,200)^2 \times 0.032}{3 (8.314)} \\\\T = 160,936.6 \ K$

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