What is the density (in kg / m^3 ) of nitrogen gas (molecular mass =28 u ) at a pressure of 2.0

What is the density (in kg / m^3 ) of nitrogen gas...

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental principle in physical chemistry and thermodynamics that relates the pressure, volume, temperature, and amount of gas through the equation PV = nRT. This law is used to understand and predict the behavior of an ideal gas under various conditions, assuming no interactions between gas molecules.

Gas Density Calculation

Gas density can be derived from the ideal gas law by expressing the mass in terms of molecular mass and the number of moles, leading to the formula ? = (MP)/(RT), where ? is the density, M is the molar mass, P is the pressure, R is the gas constant, and T is the temperature. This relation is key in linking macroscopic properties of gases to their microscopic composition.

Molecular Mass

Molecular mass, or molar mass, is a measure of the mass of one mole of a substance, usually expressed in grams per mole or kilograms per mole. It is crucial in determining the density of a gas because it quantifies the mass contained in a given number of moles, linking the microscopic scale of molecules to the macroscopic observable density.

Unit Conversion

Proper unit conversion is essential when applying formulas like the ideal gas law. This includes ensuring pressure, volume, temperature, and mass are expressed in compatible units with the gas constant. For instance, converting atmospheres to pascals or grams to kilograms is necessary to obtain the correct density in kg/m³.

Transcript

00:05 Let's consider a scenario in which we are given a few bits of valuable information.

00:10 There should be a short way.

00:15 So, we're given that the atomic mass units, or excuse me, atomic mass of diatomic nitrogen is 28 atomic mass.

00:25 We're also given that the pressure of this system is two atmospheres.

00:32 Now, we're also given that the temperature is 310 kelvin's.

00:38 So, we need to find the density of the air, kilograms of meters cubed.

01:09 So let's go ahead and find our weight to the solution.

01:13 Pardon me, that pencil, that was a pencil that just went flying.

01:18 So solution can be found using the ideal gas law.

01:25 Now the ideal gas law is a million equation that can be verified experimentally, and happens to have this ideal gas constant here, flying things in together.

01:36 Let's go ahead and take some additional information we'll need, like the ideal gas constant.

01:43 We'll keep it in metric, because everything is really nice and simple in metric.

01:50 Okay.

01:57 Let's go ahead and figure out the conversion for atmosphere.

02:00 So one atmosphere equal to 101 ,325 pascels.

02:13 I'm not going to do the factory label thing.

02:16 I think i'm just going to go out on a limb.

02:17 And say that two atmospheres equals one to 650 202 ,650 pascal.

02:30 Okay, that one other information i'd like to include before we begin working on the solution itself.

02:41 We have 28 atomic mass units corresponding to diatomic nitrogen.

02:49 Now just be aware of the fact that atomic mass units that just translates to the number of grams per 6 .02 to 2 times 10 to 23rd molecules, or atoms depending on the context, but in this context, it's molecules.

03:15 So given the ideal gas law equation, let's go ahead and play with it a little bit.

03:21 So p .e is nrt.

03:23 Dundon here, but we're almost done, or we're about halfway there.

03:27 Yeah, halfway.

03:28 There.

03:29 So let's go ahead and solve, let's go ahead and set it up this way.

03:33 So t over r t over r t.

03:37 Now remember the units we want to end up with in kilograms meters cubed.

03:46 Now you're aware of the fact that that's a little bit different from what we have going on over here.

03:55 The units you should get when you finish all of this should be the number of molecules per unit volume or per meters cubed in this case.

04:13 And i think you know where i'm going with this.

04:15 What i would suggest to you is that you end up finding the number of, excuse me, the weight of a given molecule and then use that to figure out the number of kilograms per meters cubed.

04:29 Now, just be aware of the fact that these two statements here, they're equivalent.

04:40 So we can interchange them.

04:42 So we can use all this information over here, circling it again.

04:46 And we can use that to find this value, this ratio.

04:52 We need this ratio.

04:53 We don't need any one variable.

04:55 We just need this ratio.

04:57 And this thing's units over here will work out to be exactly the same.

05:04 So we can still work with that.

05:08 So let's go ahead and plug in for pressure.

05:12 Go ahead...