Consider the following changes: a. N2(g) ⟶N2(l) b. CO(g)+H2 O(g) ⟶H2(g)+CO2(g) c. Ca3 P2(s)+6 H2

Consider the following changes: a. N2(g) ⟶N2(l) b. CO(g)+H2...

Consider the following changes: a. $\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)$ b. $\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$ c. $\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)$ d. $2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$ e. $\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)$ At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Consider the following changes:
a. $\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}(l)$
b. $\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$
c. $\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ca}(\mathrm{OH})_{2}(s)+2 \mathrm{PH}_{3}(g)$
d. $2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$
e. $\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)$
At constant temperature and pressure, in which of these changes is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done?

Key Concepts

Role of Phases in Volume Change

The phase of a substance (gas, liquid, or solid) plays a pivotal role in determining its contribution to volume changes during a reaction. Under constant temperature and pressure, gases contribute significantly to volume changes due to their high compressibility compared to liquids and solids, which typically have negligible volume change. Understanding the states of the reactants and products helps in predicting whether appreciable work is performed in a chemical process.

Pressure-Volume Work

Pressure-volume work refers to the work done when the volume of a system changes at a constant pressure. In thermodynamics, this form of work is often computed using the expression w = -P?V. The concept is crucial to understanding energy exchanges in processes where gases expand or compress, as changes in volume directly relate to the work done during the reaction.

Work Sign Convention

The sign convention for work in thermodynamics distinguishes between work done by the system and work done on the system. When a system expands, it performs work on its surroundings, resulting in a negative value of work. Conversely, when a system is compressed, the surroundings perform work on it, leading to a positive work value. This convention is essential for correctly interpreting energy transfer in chemical processes.

Change in Moles of Gas

In reactions involving gases, the difference in the number of moles of gaseous reactants and products (?n_gas) is directly linked to the change in volume under constant temperature and pressure conditions, as described by the ideal gas law. An increase in the number of gas moles generally leads to an expansion (positive ?V) and work done by the system, while a decrease results in a compression (negative ?V) and work done on the system.

Transcript

00:01 In this question, we have multiple parts here, and we're asked to figure out if there's work being done on the system or not, and whether work is going in or work is going out, or there could be a situation given these chemical reactions.

00:18 There's no work being done whatsoever.

00:20 So the first thing we need to review real quick is work in terms of pressure and volume is negative p -delta -v.

00:30 Now, we're told that pressure, pressure is constant, so that's not going to change.

00:38 And we're also given a constant temperature as well, also t.

00:45 Okay, so that means we need to look at volume.

00:48 We need to look at volume changes and if volume increases or decreases.

00:53 And so we'll look at each of these reactions and determine what's going on.

00:57 And keep in mind that when we have, again, we're using w equals, negative p delta v delta v okay so if the number of moles increases from left to right so we start off with fewer moles of gas and end up with more we know the volume will increase and therefore delta v will be positive now if we have a positive delta v will end up with work being negative that means works going out so i think i wrote that on the screen yeah so negative means work is being done by the system or work energy is going out.

01:41 And then if we have the opposite scenario where the number of gas moles decreases in this reaction, we know that we will have a decrease in volume and our delta v, which is our change in volume, will be negative.

01:55 So when we have a negative delta v times a negative p, then we know work is positive.

02:03 And then we know work is being done on the system here.

02:08 Okay.

02:09 So we'll go ahead and start with the first one.

02:12 We have nitrogen gas going to liquid nitrogen.

02:16 So we know we're essentially going down the stair step here.

02:21 Okay.

02:22 You know this is solid, liquid, and the corner we have our gas.

02:27 Okay, so we're going from the gas state to a liquid state.

02:30 Volume definitely increases as we go down the stair step.

02:34 A gaseous substance has much larger volume than its liquid substance by pretty much by definition.

02:42 So we know we're decreasing our volume.

02:46 So if we say volume is going down, we know our delta v will be negative, and therefore our work is going to be positive.

02:56 So we have work being done on the system by the surroundings.

03:02 On the system, i'll just put this off to the side.

03:06 The system by surroundings.

03:12 So that's the answer for the first one.

03:17 Now for the second one, we have carbon monoxide in the presence of water will give us hydrogen gas and carbon dioxide.

03:27 They're all gaseous substances.

03:30 And so i know i have n equals two over here.

03:33 We have one mole of gas here, of one mole of gas here, and then likewise on the other side.

03:39 So there's no change in the volume or in the number of moles of gases.

03:44 We assume there's no change in the volume as well.

03:47 And again, we have constant temperature and pressure.

03:50 That's why we can assume the volume is not changing.

03:54 So when delta v equals zero, according to our equation, w times negative p, you know, p times zero is just going to be zero...

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