Confusion about work done by a gas - Thermodynamics

[laser1]

Homework Statement

descripion

Relevant Equations

PV = nRT

This is chemistry but it's basically physics :D.

I used PV = nRT, I get V = 37.44 L. This is fine. So then I have W = P(Vfinal - Vinitial). Vinitial is zero, because there was no hydrogen gas initially. So I get 3.78 kJ. And as the gas expanded from 0 L to 37.44 L, the gas has done positive work. So the answer (I think) should be 3.78 kJ. The answer given is -3.78 kJ though... Does my logic make sense?

 

[TSny]

Your logic makes sense to me. The work done by the liberated hydrogen gas is positive.

 

[vela]

It could just be a sign convention. I've seen work defined both as $+\int p\,dV$ and $-\int p\,dV$. Which way is it defined in your class?

 

[kuruman]

Prepositions are important when dealing with "work done." Here we know that the entity doing the work is the hydrogen gas. My question on what is this work done? Furthermore, OP's suggestion that this process is isobaric from zero volume to V = 37.44 L does not make sense to me. The number of moles $n$ in the ideal gas law varies from zero to some final value. What is the pressure of zero molecules of gas? What about one or two molecules?

 

[laser1]

Prepositions are important when dealing with "work done." Here we know that the entity doing the work is the hydrogen gas. My question on what is this work done? Furthermore, OP's suggestion that this process is isobaric from zero volume to V = 37.44 L does not make sense to me. The number of moles $n$ in the ideal gas law varies from zero to some final value. What is the pressure of zero molecules of gas? What about one or two molecules?

If a molecule is exposed to the atmosphere, doesn't that mean that all of them are at atmospheric pressure?

 

[laser1]

It could just be a sign convention. I've seen work defined both as $+\int p\,dV$ and $-\int p\,dV$. Which way is it defined in your class?

+ PdV then it is work done by gas
- PdV then it is work done on gas

sign convention applies when doing calculation, but does not affect final answer in my experience of doing thermo in both chem + physics

 

[kuruman]

It could just be a sign convention. I've seen work defined both as $+\int p\,dV$ and $-\int p\,dV$. Which way is it defined in your class?

The sign convention comes in not in the definition of work but in the two forms of the first law of thermodynamics. The work done by the gas is always $dW=pdV$. It is a positive/negative quantity depending on whether the volume is increasing/decreasing. I have seen textbooks adopt two forms of the first law of thermodynamics.

1. $dU=dQ +dW$. This is what I call the work-energy theorem extension of the first law to include heat. The change in internal energy $dU$ is viewed as the change in the total kinetic energy of the gas molecules. On the other side of the equation we have two kinds of energy that cross the system boundary, the work done by external forces $dW=pdV$ and the heat $dQ$ that is added to the gas.
2. $dU=dQ -dW$. This is what I call the bank account analog of the first law. The change in internal energy $dU$ is analogous to the change in one's bank account balance. The heat added $dQ$ is analogous to account adjustments via bank actions; if positive it is fees deducted and if negative, it is interest earned. The work done by the gas $dW=pdV$ is analogous to account adjustments via actions of the account owner; if positive it is money withdrawn and if negative, it is money deposited.

 

[kuruman]

If a molecule is exposed to the atmosphere, doesn't that mean that all of them are at atmospheric pressure?

Does it? According to the kinetic interpretation of pressure, one writes $$p=\frac{1}{3}\left(\frac{N}{V}\right)m\bar{v^2}$$ where
$p=~$ the partial pressure of the hydrogen molecules
$N=~$ the number of the hydrogen molecules
$V=~$ the volume that the hydrogen molecules occupy
$m=~$ the mass of a hydrogen molecule
$\bar{v^2}=~$ the average speed-squared of the hydrogen molecules.

The 1 atm given by the problem must be the pressure of the air surrounding the beaker all contained in volume $V$. Why should the partial pressure of the released hydrogen gas be equal to 1 atm?

I think this is an ill-posed problem and I am not sure I know how to solve it or fix it.

 

[vela]

+ PdV then it is work done by gas
- PdV then it is work done on gas

The sign I wrote isn't the sign of the $p\,dV$. It's part of the definition of work, depending on which sign convention is being used. We all agree with your logic that the common-sense interpretation of the phrase work done by the gas should be a positive quantity here. But when it comes to the variable $W$, its sign depends on how it's defined, and the author of the problem may have been referring to $W$ by the phrase work done by the gas.

Anyway, I'm just pointing out it's a possible reason for the negative sign you found in the answer. It could also just be a typo (the minus sign wasn't supposed to be there or the problem statement should have said "on the gas") or a simple error.

 

[TSny]

The sign I wrote isn't the sign of the $p\,dV$. It's part of the definition of work, depending on which sign convention is being used. We all agree with your logic that the common-sense interpretation of the phrase work done by the gas should be a positive quantity here. But when it comes to the variable $W$, its sign depends on how it's defined, and the author of the problem may have been referring to $W$ by the phrase work done by the gas.

Yes. I think this is probably the source of confusion.

 

[laser1]

The sign I wrote isn't the sign of the $p\,dV$. It's part of the definition of work, depending on which sign convention is being used.

Yes, that's what I was saying. I was writing the two definitions of work, where dV = Vfinal - Vinitial.

1. Work done by the gas: W = PdV
or
2. Work done on the gas: W = -PdV

Both give the same answer in calculations.

the author of the problem may have been referring to W by the phrase work done by the gas.

But then, work done by the gas: W = PdV = P(Vf - Vi) = P(Vf) and that's still a positive answer!

Can you show the calculation that gives a negative answer?

 

[Chestermiller]

The work in the constant pressure case should be $$\Delta n\ RT$$where delta n is the change in the number of moles of gas.

 

[laser1]

Prepositions are important when dealing with "work done." Here we know that the entity doing the work is the hydrogen gas. My question on what is this work done?

The chemistry department in my uni think about work this way: they actually define w as a variable. They define w as work done. They never specify if w is work done on the gas or by the gas. They say that if w is negative, work is being done by the gas, and if w is positive, work is being done on the gas. I guess that works?

However, if work is done by the gas, w is negative. Hence they say that work done by the gas is negative. I have issues with this!

@vela

 

[Steve4Physics]

The chemistry department in my uni think about work this way: they actually define w as a variable. They define w as work done. They never specify if w is work done on the gas or by the gas. They say that if w is negative, work is being done by the gas, and if w is positive, work is being done on the gas. I guess that works?

However, if work is done by the gas, w is negative. Hence they say that work done by the gas is negative. I have issues with this!

It sounds like your chemistry department is defining W as the work done on a system by an external force.

E.g. suppose the system (S) is some gas enclosed in a cylinder+piston. When the piston is pushed in, the work done on S (by some external force) is positive, so the value of W is positive. But in this case the work done by the gas (on the piston) is a negative,quantity.

When defining W, ‘what is doing work on what’ needs to be clear/unambiguous.

 

[256bits]

However, if work is done by the gas, w is negative. Hence they say that work done by the gas is negative. I have issues with this!

Are you referring to the work being done on the system, or on the environment, for the proper sign convention to follow? Which one are we interested in here - the system or the environment?

For the system perspective, which work( a category of energy flow ) - negative or positive - increases / decreases the system energy.

 

[DrBanana]

Does it? According to the kinetic interpretation of pressure, one writes $$p=\frac{1}{3}\left(\frac{N}{V}\right)m\bar{v^2}$$ where
$p=~$ the partial pressure of the hydrogen molecules
$N=~$ the number of the hydrogen molecules
$V=~$ the volume that the hydrogen molecules occupy
$m=~$ the mass of a hydrogen molecule
$\bar{v^2}=~$ the average speed-squared of the hydrogen molecules.

The 1 atm given by the problem must be the pressure of the air surrounding the beaker all contained in volume $V$. Why should the partial pressure of the released hydrogen gas be equal to 1 atm?

I think this is an ill-posed problem and I am not sure I know how to solve it or fix it.

I also want to know why we can use the pressure of the atmosphere. Is it because of the following reasoning:
If W, F and delta s are work, force (on the atmosphere) and displacement, respectively, then :
$W=F\Delta s$
$W=\frac{F}{A} A \Delta s$, $A$ is some cross sectional area
$W=P \Delta V$
Crucially, this $P$ comes from the force being exerted on the atmosphere, moreover, it is connected to the minimum force needed to push against the atmosphere. Converting that to pressure, that becomes $1 atm$ as given in the problem.

I might be completely wrong though so I'd like some feedback.