Is My Solution for Part D Logically Correct?

[EnricoHendro]

Homework Statement

A 4.00-L sample of a diatomic ideal gas with specific heat ratio 1.40, confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. (a) Determine the volume of the gas at the end of the adiabatic expansion. (b) Find the temperature of the gas at the start of the adiabatic expansion. (c) Find the temperature at the end of the cycle. (d) What was the net work done on the gas for this cycle?

Relevant Equations

PV=nRT
P(V^y) = Constant

Hello there, is my solution for part d logically correct? Here is my attempt at the solution :
Part a :

where : P1 = 3P2

Part b :
Since P1=3P2, therefore, T1=3T, where T=300K. Thus, T1=900K

Part c :
Because the final pressure at the end of the cycle is exactly the same as the pressure at the very beginning of the cycle and the volume at the end of the cycle is equal to the very beginning of the cycle, then, T3 = T = 300K, (with T3 = Temperature at the end of the cycle)

Part d :

W1 = 0 (Because the first process is isovolumetric)

(since the process is adiabatic, where Q=0)

(isobaric process)
since :
n is constant assuming no leak and no additional gas during the whole process

and

With :

Therefore :

Where,
Vf=V2
Vi=V1
T1=900K

My textbook's answer in part d is also -336Joule, but the method in the solution manual is different than mine.
p.s. The volume is in cubic meter

Thank you

 

[kumusta]

Can you not show us a pV diagram of the problem? Thanks.

 

[Chestermiller]

Homework Statement:: A 4.00-L sample of a diatomic ideal gas with specific heat ratio 1.40, confined to a cylinder, is carried through a closed cycle. The gas is initially at 1.00 atm and 300 K. First, its pressure is tripled under constant volume. Then, it expands adiabatically to its original pressure. Finally, the gas is compressed isobarically to its original volume. (a) Determine the volume of the gas at the end of the adiabatic expansion. (b) Find the temperature of the gas at the start of the adiabatic expansion. (c) Find the temperature at the end of the cycle. (d) What was the net work done on the gas for this cycle?
Relevant Equations:: PV=nRT
P(V^y) = Constant

Hello there, is my solution for part d logically correct? Here is my attempt at the solution :
Part a :
View attachment 289142
View attachment 289143
where : P1 = 3P2

Part b :
Since P1=3P2, therefore, T1=3T, where T=300K. Thus, T1=900K

Part c :
Because the final pressure at the end of the cycle is exactly the same as the pressure at the very beginning of the cycle and the volume at the end of the cycle is equal to the very beginning of the cycle, then, T3 = T = 300K, (with T3 = Temperature at the end of the cycle)

Part d :
View attachment 289145
W1 = 0 (Because the first process is isovolumetric)
View attachment 289146 (since the process is adiabatic, where Q=0)
View attachment 289147 (isobaric process)
since :
n is constant assuming no leak and no additional gas during the whole process
View attachment 289149
and
View attachment 289150
With :
View attachment 289151
View attachment 289152
Therefore :
View attachment 289153
Where,
Vf=V2
Vi=V1
T1=900K

My textbook's answer in part d is also -336Joule, but the method in the solution manual is different than mine.
p.s. The volume is in cubic meter

Thank you

It looks like you have a sign error in your last equation (in front of the last term). Please show us what your book did. Did they work with the heats instead of the works? Don't forget that, for the entire cycle, Q = W.

 

[EnricoHendro]

It looks like you have a sign error in your last equation (in front of the last term). Please show us what your book did. Did they work with the heats instead of the works? Don't forget that, for the entire cycle, Q = W.

In the book, it states W = -Q since final temperature after the whole cycle equals to the initial temperature (300K), therefore, the change in internal energy is zero. Using 1st law of thermodynamics, W=-Q. The book evaluates the Q for each cycle, and then using change in internal energy = 0, W=-Q

Can you not show us a pV diagram of the problem? Thanks.

oh, sorry, I forgot to include the diagram. Here it is. Mine's the one on whiteboard, it's a very rough sketch, and I will also include the diagram from the book

 

[EnricoHendro]

It looks like you have a sign error in your last equation (in front of the last term). Please show us what your book did. Did they work with the heats instead of the works? Don't forget that, for the entire cycle, Q = W.

here is the method that my book used

 

[Chestermiller]

here is the method that my book used View attachment 289160View attachment 289161

Oh. The form of the 1st law I work with is $\Delta U=Q-W$, where W is the work done by the system on the surroundings. Apparently you use $\Delta U=Q+W$, where W is the work done by the surroundings on the system.

Either of the two methods (yours or your book's) is acceptable. I would have used their method, because it seems easier (to me).

 

[EnricoHendro]

Oh. The form of the 1st law I work with is $\Delta U=Q-W$, where W is the work done by the system on the surroundings. Apparently you use $\Delta U=Q+W$, where W is the work done by the surroundings on the system.

Either of the two methods (yours or your book's) is acceptable. I would have used their method, because it seems easier (to me).

oh I see. Yes, my textbook taught me that the change in internal energy is Q+W.

I see. Thank you for your answer

 

[kumusta]

I just want to make some clarifications regarding the two forms of the first law that Chestermiller mentioned: $\Delta U = Q \pm W$

Oh. The form of the 1st law I work with is $\Delta U=Q-W$, where W is the work done by the system on the surroundings. Apparently you use $\Delta U=Q+W$, where W is the work done by the surroundings on the system ... Either of the two methods ... is acceptable ...

In both of them, $Q \gt 0$ for the heat absorbed by (added to) the system, while $Q \lt 0$ for the heat given off, removed from the system. They differ in the sign convention for $W~$though.
$~\Delta U = Q - W~\Rightarrow~\begin{cases}\begin{align} &W\gt~0~ \text { for work done by system (volume expansion)}\dots \nonumber \\ &W\lt~0~ \text { for work done on system (volume compresion)}\dots\nonumber\end{align}\end{cases}$
To understand the above sign convention, you have to remember than the internal energy$~\Delta U~$depends on the temperature only. Since volume compression generally leads to rise in temperature $-$ just think of a bicycle tire slightly warming up when too much air is pumped into it $-$ making $W\lt~0~$in the first form of the first law given above adds more heat to $Q$ causing$~\Delta U~$to increase. On the other hand, with $W\gt~0~$during volume expansion (removing some air inside the tire to give more space to the gas particles inside), heat is removed instead from $Q$ causing $\Delta U~$to decrease which then leads to a drop in temperature of the tire.
The sign convention for $W$ in the second form of the first law is just opposite to that in the first form.
$~\Delta U = Q + W~\Rightarrow~\begin{cases}\begin{align} &W\lt~0~ \text { for work done by system (volume expansion)}\dots \nonumber \\ &W\gt~0~ \text { for work done on system (volume compresion)}\dots\nonumber\end{align}\end{cases}$