At a certain point in a coal gasification process, a sample...

At a certain point in a coal gasification process, a sample of the gas is taken and stored in a $1 \mathrm{L}$ cylinder. An analysis of the mixture yields the following results: $$\begin{array}{lllll} \hline \text { Component } & \mathbf{H}_{2} & \mathrm{CO} & \mathrm{CO}_{2} & \mathrm{N}_{2} \\ \hline \text { Percent by mass } & 2 & 45 & 28 & 25 \\ \hline \end{array}$$ Determine the mole fractions and total mass in the cylinder at $100 \mathrm{kPa}$ and $20^{\circ} \mathrm{C}$. How much heat must be transferred to heat the sample at constant volume from the initial state to $100^{\circ} \mathrm{C} ?$

At a certain point in a coal gasification process, a sample of the gas is taken and stored in a $1 \mathrm{L}$ cylinder. An analysis of the mixture yields the following results:
$$\begin{array}{lllll}
\hline \text { Component } & \mathbf{H}_{2} & \mathrm{CO} & \mathrm{CO}_{2} & \mathrm{N}_{2} \\
\hline \text { Percent by mass } & 2 & 45 & 28 & 25 \\
\hline
\end{array}$$
Determine the mole fractions and total mass in the cylinder at $100 \mathrm{kPa}$ and $20^{\circ} \mathrm{C}$. How much heat must be transferred to heat the sample at constant volume from the initial state to $100^{\circ} \mathrm{C} ?$

Key Concepts

Specific Heat at Constant Volume

The specific heat capacity at constant volume (Cv) indicates the amount of heat required to raise the temperature of a unit amount of substance by one degree at constant volume. This property is used to calculate the energy needed to heat the gas mixture from its initial temperature to the desired final temperature.

Constant Volume Process

A constant volume process implies that the volume remains fixed during the process. In such a process, any heat added is used solely to increase the internal energy of the gas, as no work is done by expansion. This concept is essential when determining the amount of heat required to raise the temperature of the gas.

Mass Percent Composition

Mass percent composition represents the proportion of each component in a mixture by mass. By combining this with the molar masses of the components, one can determine the mass and number of moles of each individual gas, which is crucial for establishing the overall composition in both mass and mole terms.

Mole Fraction

Mole fraction is the ratio of the number of moles of a given component to the total number of moles in the mixture. It is a key concept for expressing the composition of a gas mixture in a way that is independent of the physical conditions, and is particularly useful for calculations involving partial pressures and chemical reactions.

Ideal Gas Law

The ideal gas law (PV = nRT) is a fundamental relation used to connect pressure, volume, temperature, and the number of moles of a gas. In this context, it allows determination of the total number of moles of the gas mixture in the cylinder, using the known volume, temperature, and pressure.

Transcript

00:01 Hi, everybody.

00:02 So this one's asking us, how much heat must be transferred to the heat sample of constant volume in this state to r to 3 celsius, okay? so for this, we have pv equals mr mix times t.

00:25 And we need to define the r mix first.

00:29 Okay? so for that, we're going to do what we're going to do is do 0 .02 for each 2c2 and i'll write this out in just a second so let me rate out these numbers you got 4 .1 243 for the mi okay and so this is your ci the mi here of h2.

01:05 And now we have 0 .45 times 0 .2968.

01:16 And that is your co.

01:19 And i'll write it down just below in just a second.

01:21 So we have 0 .28 times it was 0 .8.

01:32 Make sure i get the right number.

01:35 Number times 0 .189, okay, plus 0 .25 times 0 .29.

01:55 Okay? and let me just, um, put that in here, i want this here.

02:02 0 .2968.

02:06 Here, this one is your co.

02:10 Here, this is your c -o.

02:12 Here, this is, your co2.

02:18 And this one is your nitrogen, too, okay? and you can find all these numbers in the chart, in your book, okay? and so that is going to give us 0 .3 -431 kilojoules, kilograms.

02:41 You're not going to have to memorize this.

02:43 It's okay.

02:43 If you don't have to memorize this, it's okay, if you don't quite know, you can look in the book.

02:46 Your professors not can require, hopefully not going to require you to number 1 ,000 numbers, okay? physics is a really relaxed field, and that's why i love it...

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