A calculate the rms speed of an oxygen gas molecule o2 at...

a) Calculate the rms speed of an oxygen gas molecule (O2) at 31.0°C. b) At the same temperature, which molecule travels faster, O2 or N2? c) Using your answer to question 8b above, calculate how much faster? d) Standard Temperature is defined to be 273.15K. What is the average translational kinetic energy of a single molecule of an ideal gas at Standard Temperature? e) One mole (mol) of any substance consists of 6.022 x 10^23 molecules (Avogadro's number). What is the translational kinetic energy of 12.00 mol of an ideal gas at 300.0K? f) A sealed syringe contains 10 × 10^-6 m^3 of air at 1 × 10^5 Pa. The plunger is pushed until the volume of trapped air is 4 x 10^-6 m^3. If there is no change in temperature, what is the new pressure of the gas? g) A car tyre contains air at 1.25 × 10^5 Pa when at a temperature of 27°C. Once the car has been running for a while, the temperature of the air in the tyre rises to 42°C. If the volume of the tyre does not change, what is the new pressure of the air in the tyre?

Transcript

00:01 Okay, so this post has quite a few parts.

00:04 So let's start off with part a.

00:07 So we have v of rms is equal to the square root of 3rt over m, where r is equal to 8 .314, t is equal to 31 degrees celsius or 304 kelvin's.

00:41 And m is molar mass.

00:48 So that will be the 32 grams per mole.

00:54 And that is also equal to 0 .032.

00:58 So now we plug what we know into this equation.

01:02 So then we have equals the square root.

01:06 Of, it's going to have to move this down, equals a square root of 3 times 8 .314 times 304 divided by the molar mass of 0 .032, and that gives you 486 .77 meters per second squared.

01:35 So now we need part b.

01:50 So moving down for part b, we have the kinetic energy of the atoms of molecules of the gas, so k .e is equal to one half mv squared.

02:04 So if two masses of nitrogen and oxygen are at the same temperature, that means the kinetic energy value is equal to one half mv2, vn2.

02:21 Sorry, this guy right here is supposed to be an n as well.

02:26 So mn2 vn2 squared is equal to 1 half m02 v02 squared.

02:41 So plugging stuff that we know into this equation, here we have n2 has a mass of 28 amus while o2 has a mass of 32 amus so that means that 1 1 .5 28 v .n2 squared is equal to 1 1 1�32 v02 squared so then vn2 over v02 equal to the square root of 32 over 28.

03:22 So then vn2 over v02 is equal to 1 .06.

03:31 So that is proved that n2 travels 1 .06 times greater than 02.

03:50 So then next, sorry, that was part c what i just said.

04:02 1 .06 times faster than 02.

04:12 So c basically just wanted us to put that in words.

04:15 So d, the average translational energy of a molecule is e equals 3kt over 2, where k is a constant, t is obviously temperature, and that is equal to 273 .15 kelvin's.

04:41 So then we have e is equal to three halves times 273 .15 times 1 .38 times 10 to the negative 23 jules.

04:54 And so e will then equal 565 .4 times 10 to negative 23 joules.

05:02 And that would be your answer for part d.

05:04 Moving on to e.

05:08 So it is known that relation between kinetic energy and number of moles and temperature is as follows.

05:18 So we have kinetic energy is equal to three halves in rt.

05:23 So n is 12 moles...