What is the change in internal energy after two processes?

[Callmelucky]

Homework Statement

Can someone please help me understand what I am doing wrong

Relevant Equations

PV=nRT, W=p*deltaV, deltaU=Q-W, Q=nCvdeltaT

So the question goes like this: find change in internal energy in process 1->2 using diagram. And offered solutions a)-400J b)400J c)600J d)800J.
First I found T1 and T2 using (P*V)/T=R and got T1=24K and T2=72K. Then I found n(number of moles) using PV=nRT and got n1=1mol, n2=1mol. Then I calculated work using W=p*deltaV and got W=200J. After that, I did Q=deltaU -> Q=nCvdeltaT and got Q=598.6J rounded that to 600J, and subtracted that from work because deltaU=Q-W and got deltaU=400J. But the solution is supposed to be 600J.

So can someone please tell me where did I make a mistake.

Thank you.

 

[Callmelucky]

Edit: I subtracted work from heat not heat from work

 

[kuruman]

Homework Statement:: Can someone please help me understand what I am doing wrong
Relevant Equations:: PV=nRT, W=p*deltaV, deltaU=Q-W, Q=nCvdeltaT

So the question goes like this: find change in internal energy in process 1->2 using diagram. And offered solutions a)-400J b)400J c)600J d)800J.
First I found T1 and T2 using (P*V)/T=R and got T1=24K and T2=72K. Then I found n(number of moles) using PV=nRT and got n1=1mol, n2=1mol. Then I calculated work using W=p*deltaV and got W=200J. After that, I did Q=deltaU -> Q=nCvdeltaT and got Q=598.6J rounded that to 600J, and subtracted that from work because deltaU=Q-W and got deltaU=400J. But the solution is supposed to be 600J.

So can someone please tell me where did I make a mistake.

Thank you.

The work done by the gas from 1 to 2 is more than 200 J. The 200 J that you found is the work done when the gas expands at constant pressure from 1 L to 2 L. This is not the case here. Remember that the work done by the gas is the area under the curve from 1 to 2. Here, the "curve" is a trapezoid.

 

[Callmelucky]

The work done by the gas from 1 to 2 is more than 200 J. The 200 J that you found is the work done when the gas expands at constant pressure from 1 L to 2 L. This is not the case here. Remember that the work done by the gas is the area under the curve from 1 to 2. Here, the "curve" is a trapezoid.

Thank you for answering.
That is true. It is slightly bigger work(50J more, so total =250J) but that option is not in the answer

 

[kuruman]

Since ΔU does not depend on the path, it is OK to consider a convenient path from 1 to 2 like an isobaric + isochoric process. However, if you do that you need to account correctly for the heat added to and the work done by the gas because the two do depend on the path.

Can you post the statment of the problem exactly as it was given to you? I suspect that there is more to it than you have mentioned. By the way, your calculation for the number of moles is flawed.

First I found T1 and T2 using (P*V)/T=R and got T1=24K and T2=72K

You used the ideal gas law in which you set $n=1$ to find the temperature. Then you used the ideal gas law again to find $n=1$ which is going around in a circle. You cannot get two unknown quantities from one equation.

 

[Callmelucky]

Since ΔU does not depend on the path, it is OK to consider a convenient path from 1 to 2 like an isobaric + isochoric process. However, if you do that you need to account correctly for the heat added to and the work done by the gas because the two do depend on the path.

Can you post the statment of the problem exactly as it was given to you? I suspect that there is more to it than you have mentioned. By the way, your calculation for the number of moles is flawed.

You used the ideal gas law in which you set n=1 to find the temperature. Then you used the ideal gas law to find n=1 which is going around in a circle. You cannot get two unknown quantities from one equation.

The problem goes: what is the value of internal energy change of ideal gas in process 1->2 as shown in picture?

Just realized that you were right about going in circle, that is why I have deleted picture of textbook page

 

[kuruman]

I don't understand Croatian. You have to read this carefully and explain to me the meaning of the equation that you circled $$\frac{pV}{T}=R$$ and then how this is converted to $$\frac{pV}{T}=nR$$It seems that the first equation is some kind of preliminary form of the ideal gas law that is fully expressed in the second equation. This means that you cannot use these as two independent equations which is what you did.

On edit: The textbook page to which this post refers has been deleted by the OP.

 

[Callmelucky]

I don't understand Croatian. You have to read this carefully and explain to me the meaning of the equation that you circled $$\frac{pV}{T}=R$$ and then how this is converted to $$\frac{pV}{T}=nR$$It seems that the first equation is some kind of preliminary form of the ideal gas law that is fully expressed in the second equation. This means that you cannot use these as two independent equations which is what you did.

Problem was that molar volume was not explained in the textbook so I thought that n stands for mole and it turned out that n stood for the number of moles. I didn't know that molar volume was used as V in pV/T=R.

Edit: still, I assumed that number of moles was 1, okay, but it doesn't say in the task what number of moles is.
Edit 2: can you tell me how would you solve it? You don't have to write steps in equation as above just tell me what to do.

 

[Steve4Physics]

Hi @Callmelucky. You don’t need the number of moles or the temperatures. Try this…

For $n$ moles of an ideal gas, $PV = nRT$
The internal energy of an ideal gas is $U = \frac 32 nRT$
What is the formula for $U$ in terms of $P$ and $V$?

Now can use your formula to answer the question?

 

[Callmelucky]

Ohh yes :D U=3/2pV. Calculate the 1st, 2nd, subtract and that's it. Thank you @Steve4Physics

 

[kuruman]

The internal energy of an ideal gas is $U = \frac {3}{2} nRT$

Assuming, of course, that the gas is monatomic.