Solving a Monatomic Ideal Gas Expansion Problem

• VincentweZu
In summary: W=+3600J and ΔU=+5400J.If W is work done by the system, then W=-3600J and ΔU=-5400J. That would effect the answer in iii) so the Q is +9000J.Correct.Sorry but I'd just like to clarify the last line. Do you mean that, when heat is gained in an isobaric process then the gas is expanding and doing work, whereas when heat is lost in an isobaric process then the gas is shrinking and work is being done one it? And so technically heat can be lost or gained in an isobaric process?Yes, that's correct. Heat can be lost or
VincentweZu

Homework Statement

the diagram above of pressure P versus volume V shows the expansion of 2.0 moles of a monatomic ideal gas from state A to state B. As shown in the diagram PA = PB = 600 N/m2 , VA = 3.0 m3 and VB = 9.0 m3.

a)
i) Calculate the work done by the gas as it expands
ii) Calculate the change in internal energy of the gas as it expands.
iii) Calculate the heat added to or removed from the gas during this expansion.

Homework Equations

W = PΔV
U = $\frac{3}{2}$nRT
PV = nRT

The Attempt at a Solution

a)
i) W = PΔV
W = 600(9.0 - 3.0)
W = - 3600J
I put negative to indicate work done by system

ii) Here is where I have problems, I'm not sure how to calculate internal energy without given temperature. Here is my failed attempt.

UA = $\frac{3}{2}$nRTA
Since PV = nRT
UA = $\frac{3}{2}$PAVA

ΔU = UB - UA
= $\frac{3}{2}$PBVB - $\frac{3}{2}$PAVA
Since PA = PB = P
= $\frac{3}{2}$P(VB - VA)
= $\frac{3}{2}$PΔV
= $\frac{3}{2}$W
= $\frac{3}{2}$(-3600)
= -5400J

I am not sure that this is correct because I don't know if I am allowed to rearrange the equations like that, although it is correct algebraically, I'm not sure if it is correct physically.
I am unsure of whether the change in internal energy is supposed to be negative or positive, negative makes sense to me because work is being done by the gas and thus the internal energy should decrease.

iii) ΔU = Q + W
-5400 = Q -3600
Q = -1800J

Now I know there is something wrong since heat cannot be lost by the gas in an isobaric process.

VincentweZu said:

The Attempt at a Solution

a)
i) W = PΔV
W = 600(9.0 - 3.0)
W = - 3600J
I put negative to indicate work done by system
Looks good. Strictly speaking, the answer to the question as asked is +3600 J, even though you are correct that W is -3600 J.

ii) Here is where I have problems, I'm not sure how to calculate internal energy without given temperature.
Temperature could be calculated from n, P, and V. But your method is simpler and better.
Here is my failed attempt.
UA = $\frac{3}{2}$nRTA
Since PV = nRT
UA = $\frac{3}{2}$PAVA

ΔU = UB - UA
= $\frac{3}{2}$PBVB - $\frac{3}{2}$PAVA
Since PA = PB = P
= $\frac{3}{2}$P(VB - VA)
= $\frac{3}{2}$PΔV
= $\frac{3}{2}$W
Is PΔV equal to W or -W?
= $\frac{3}{2}$(-3600)
= -5400J

I am not sure that this is correct because I don't know if I am allowed to rearrange the equations like that, although it is correct algebraically, I'm not sure if it is correct physically.
I am unsure of whether the change in internal energy is supposed to be negative or positive, negative makes sense to me because work is being done by the gas and thus the internal energy should decrease.
The problem with this argument is that you don't know Q (yet). It is possible that Q could offset the effect of the work done, and reverse the sign of ΔU.

A better way to check the sign of ΔU is to think about whether the temperature increases or decreases during the process. You don't have to actually calculate T, just apply the ideal gas law to the case where P is constant and V increases.

iii) ΔU = Q + W
-5400 = Q -3600
Q = -1800J

Now I know there is something wrong since heat cannot be lost by the gas in an isobaric process.
Yes it can. Or it could be gained. Q=0 for adiabatic processes, not isobaric.

Hope this helps.

Redbelly98 said:
Looks good. Strictly speaking, the answer to the question as asked is +3600 J, even though you are correct that W is -3600 J.

Temperature could be calculated from n, P, and V. But your method is simpler and better.

Is PΔV equal to W or -W?

The problem with this argument is that you don't know Q (yet). It is possible that Q could offset the effect of the work done, and reverse the sign of ΔU.

A better way to check the sign of ΔU is to think about whether the temperature increases or decreases during the process. You don't have to actually calculate T, just apply the ideal gas law to the case where P is constant and V increases.

Yes it can. Or it could be gained. Q=0 for adiabatic processes, not isobaric.

Hope this helps.

Thanks it helps alot.

PΔV is equal to W. I guess ii) is wrong since numerically the value for W is +3600J not -3600J
Thus the ΔU should be +5400J not -5400J

That would effect the answer in iii) so the Q is +9000J.

Sorry but I'd just like to clarify the last line.
Do you mean that, when heat is gained in an isobaric process then the gas is expanding and doing work, whereas when heat is lost in an isobaric process then the gas is shrinking and work is being done one it? And so technically heat can be lost or gained in an isobaric process?

VincentweZu said:
Thanks it helps alot.

PΔV is equal to W. I guess ii) is wrong since numerically the value for W is +3600J not -3600J
Thus the ΔU should be +5400J not -5400J
You're correct that ΔU is +5400J.

That being said, we should clarify what W is. It can be defined as either work done on the system, or work done by the system. Both definitions get used in practice, and it's good to be clear on which definition is used in your present course -- and to use that definition consistently.

If W is work done on the system, then
• W = -PΔV = -3600 J
• ΔU = Q + W

If W is work done by the system, then
• W = +PΔV = +3600 J
• ΔU = Q - W

(Either way, you subtract 3600 J from Q to get ΔU.)

That would effect the answer in iii) so the Q is +9000J.
Yes.
For this type of problem, I like to think about what is likely happening in a real device. For this problem, I am imagining a gas in a sealed container with a moveable piston. It appears that the gas is being actively heated -- meaning it is heated by a flame or some other means. As it heats, it expands, as we know gases tend to do when heated. By expanding, some of the energy gained by heating is lost via the gas doing work on the surroundings. So ΔU is still positive -- as it must be if the temperature increased -- but less than the heat added to the gas.

(Sorry if my rambling here isn't clear.)

Sorry but I'd just like to clarify the last line.
Do you mean that, when heat is gained in an isobaric process then the gas is expanding and doing work, whereas when heat is lost in an isobaric process then the gas is shrinking and work is being done one it? And so technically heat can be lost or gained in an isobaric process?
Yes.
You're class may or may not yet have discussed processes where Q=0. For an ideal gas, it looks like processes DA and BC in this diagram:

Redbelly98 said:
You're correct that ΔU is +5400J.

That being said, we should clarify what W is. It can be defined as either work done on the system, or work done by the system. Both definitions get used in practice, and it's good to be clear on which definition is used in your present course -- and to use that definition consistently.

If W is work done on the system, then
• W = -PΔV = -3600 J
• ΔU = Q + W

If W is work done by the system, then
• W = +PΔV = +3600 J
• ΔU = Q - W

(Either way, you subtract 3600 J from Q to get ΔU.)

Yes.
For this type of problem, I like to think about what is likely happening in a real device. For this problem, I am imagining a gas in a sealed container with a moveable piston. It appears that the gas is being actively heated -- meaning it is heated by a flame or some other means. As it heats, it expands, as we know gases tend to do when heated. By expanding, some of the energy gained by heating is lost via the gas doing work on the surroundings. So ΔU is still positive -- as it must be if the temperature increased -- but less than the heat added to the gas.

(Sorry if my rambling here isn't clear.)

Yes.
You're class may or may not yet have discussed processes where Q=0. For an ideal gas, it looks like processes DA and BC in this diagram:

I'd just like to thank you, you've cleared up so much!

1. What is a monatomic ideal gas?

A monatomic ideal gas is a theoretical model of a gas that consists of particles that do not interact with each other and have no internal structure. It is often used in physics and chemistry to simplify calculations and understand the behavior of gases.

2. How is an ideal gas expansion problem solved?

An ideal gas expansion problem can be solved using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. The problem may also involve using the first law of thermodynamics and the equation for work, W = -PΔV.

3. What are the assumptions made when solving a monatomic ideal gas expansion problem?

The main assumptions made when solving a monatomic ideal gas expansion problem are that the gas particles have no volume, there is no intermolecular attraction or repulsion between particles, and the collisions between particles and the walls of the container are perfectly elastic.

4. How is the final temperature of an ideal gas calculated during expansion?

The final temperature of an ideal gas during expansion can be calculated using the equation Tf = Ti(Vf/Vi)^2, where Tf and Ti are the final and initial temperatures, and Vf and Vi are the final and initial volumes, respectively. This equation is derived from the ideal gas law and assumes that the pressure and number of moles remain constant.

5. What are some real-life applications of solving monatomic ideal gas expansion problems?

Solving monatomic ideal gas expansion problems can help in the design and analysis of various systems, such as car engines, refrigerators, and air conditioning units. It can also be used in understanding the behavior of gases in chemical reactions and industrial processes.

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