At what temperature the rms velocity of the a molecular...

At what temperature the rms velocity of the (a) molecular hydrogen (1.67x10 27 kg) and oxygen (2.66\times 10-20 kg) molecular are equal to the Earth's escape velocity (Earth's radius = 6.37x10\deg m and g=9.8 m/s2)? (b) At what temperature does the rms velocity reach the rms value of molecular hydrogen and molecular oxygen are equal to the escape velocity of the Moon (where the radius of the moon is 1.74x 10\deg m and the gravitational acceleration at the surface has magnitude of 0.16g)? (c) Considering the responses to parts (a) and (b), should there be a lot of hydrogen and oxygen at high levels in the Earth's upper atmosphere, where the temperature is about 1000 K?

Transcript

00:01 In this question we will be solving for the root mean square speed of the molecules in a gas.

00:07 So root mean square's speed of molecules in a gas is given by vrms is equal to square root of 3rt over m, where t is the temperature and m is the molar mass of the gas.

00:25 The speed required for escape from earth's gravitational pool, is v equals square root of 2g re where g is the acceleration due to gravity and re is equal to 6 .37 times 10 to the 6 meters which is the radius of the earth so to derive this expression we take the zero of gravitational potential energy to be at infinity then the gravitational potential energy of a part with mass m that earth's surface is u equals negative g m m over r e squared is equal to negative m g r e squared is equal to negative m g re where g equals g m over r e squared was used so if b is the speed of the particle then its original energy is e equals negative m g r e plus one half m v squared and if the particle is just able to travel far away its kinetic energy must tend towards zero as it advanced as its distance from earth becomes large without bound this means e equals zero and v equals square root of two g r e we equate the expressions for the speeds to obtain square root of 3 r t divided by m is equal to square root of 2 g rea the solution for t is t equals 2 g r m over 3r so therefore for letter a the molar mass of hydrogen is 2 .02 times 10 to the negative 3 kilogram per mole so temperature is equal to or 2 times 9 .8 meters per second square times 6 .37 times 10 to the 6 meters times 2 .02 times 10 to the negative 3 kilogram per mole divided by 3 times 8 .31 joules per mole.

03:07 Mole kelvin since we are asked to solve for the temperature therefore t is equal to 1 times 10 to the 4 kelvin next for letter b using the same equation we have molar mass of oxygen that we will use so t equals 2 times 9 .8 means per second squared 6 .3 7 times 10 to the 6 meters times 3 .2 sorry 32 32 times 10 to the 3 kilogram per mole divided by 3 times 8 .31 joules per mole kelvin will give us the temperature of 1 .6 times 10 to the 10 to the 5 kelvin next for letter c we have now that now t is equal to 2 gm r m over 3 r where r m is equal to 1 .74 times 10 to the 6 meters which is the radius of the moon and gm is .16 grams is the acceleration due to gravity at the moon surface for hydrogen.

05:15 So the temperature is equal to t equals 2 times .16 times 9 .8 meters per second squared times 1 .74 times 10 to the 6 meters times 2 .02 times 10 to the negative 3 kilogram per mole divided by 3 times 8 .31 choles per mole kelvin...

Question

Please give Ace some feedback