Weighing of masses gives a mixture at 60^∘ C 225 kPa with 0.5 kg O2, 1.5 kg N2, and 0.5 kg CH4 .

Weighing of masses gives a mixture at 60^∘ C 225 kPa with...

Key Concepts

Mass-Based Specific Volume

Mass-based specific volume is defined as the volume occupied per unit mass of a substance (or mixture). In the context of gas mixtures, it can be derived by combining the ideal gas law with the overall mass of the mixture, converting molar properties into quantities expressed per unit mass.

Mixture Molecular Weight Determination

The molecular weight of a gas mixture is calculated as the mole-fraction-weighted average of the individual molecular weights of the components. This concept allows for the computation of weighted properties, such as specific volume, and is crucial when relating the mass basis to molar basis in thermodynamic calculations.

Total Volume Calculation

Total volume calculation for a gas mixture involves applying the ideal gas law to the mixture. By using the total number of moles calculated from the mass and mixture molecular weight and knowing the temperature and pressure, the overall volume can be readily determined.

Dalton's Law of Partial Pressures

Dalton's law states that in a mixture of non-reacting gases the total pressure exerted is equal to the sum of the partial pressures of the individual gases. Each gas's partial pressure is proportional to its mole fraction in the mixture, and this principle is essential for determining individual gas pressures in mixtures.

Ideal Gas Law

The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas through the equation PV = nRT. It is a fundamental relation used to calculate any one of these state properties when the others are known and is especially useful when dealing with gases under conditions where they behave ideally.

Mole Fraction Calculation

Mole fraction is the ratio of the number of moles of a given component to the total number of moles in a mixture. It is a key parameter when applying Dalton’s law, as it directly relates to the partial pressures of the components in a gas mixture.

Transcript

00:01 Hi, everybody.

00:01 So this one is asking us, find the partial pressures of each component, the mixture specific volume, and mixture molecular weight and the total volume.

00:13 Okay.

00:15 So now what we can do is we can find the end here, okay? and the end total, okay? the number of molecules total.

00:33 And so we can simply write it kind of like this, okay? and so what that means is we have 0 .5, and then we have 31 .99 plus 1 .5 plus over 28 .013.

00:55 And if you're curious when i'm getting these numbers, what it is is i'll put it right under.

01:03 Underneath and green.

01:06 This is the 02.

01:10 This is the n2 here.

01:14 And the next one is 0 .05 over 16 .043.

01:28 And this one is the ch4.

01:29 Okay.

01:29 And this one is the ch4.

01:33 Okay? and what i'm going to do is just simplify it a little more.

01:41 So we have 0 .01 -6255 plus 0 .5355 plus 0 .0 .3156.

01:58 You don't have to do that.

02:00 I just simplified it a little bit more into decimals just so it's easy for you guys to see.

02:07 It and that's when we go 0 .103 okay and again you can find this number through the pr table okay so now we have the y02 and that's going to be 0 .01562 so this uses and i'll write it out in just a second so we have 0 .103 so what this is uses is and what i'll put here is we have i am i divided by the end total and from that we have 0 .1578 okay and now we have y in 2 equals 0 .157 8 okay and now we have y in 2 equals 0 .5 35355 divided by 0 .103.

03:16 And we got 0 .53386.

03:22 And now we have y -ch4 equals 0 .03116, divided by 0 .103 equals 0 .303 equals 0 .3.

03:38 One, zero, okay? we're not done.

03:41 This is a really long question, so let's bear with me.

03:46 So now we have a p -a equals y -a pressure.

03:52 Okay, and so that's our equation here.

03:55 So we have for the oxygen, we have your y -0 -2 and your feet.

04:01 Okay, so we have 0 .1, what was it, 0 .155...