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Imagine some helium in a cylinder with an initial volume of 1 liter ant initial pressure of 1 atm. Somehow the helium is made to expand to a final volume of 3 liters, in such a way that its pressure rises in direct proportion to its volume. Calculate the work done on the gas during this process, assuming there are no other types of work being done.

a) Calculate the work done by on the gas during this process, assuming that there are no other types of work being done.

b) Calculate the change in the helium's energy content during this process.

c) Calculate the amount of heat added to or removed from helium during this process.

So a) has already been dealt with here.

b) The total change in energy is given by $$\Delta U = Q+W\tag1$$ where $Q$ is heat energy and $W$ is work done on the system. $W=-405 \ \text{J}$ is already given by the answer to question a), but how do I get the heat energy $Q?$

My first thought was that we can first compute $T_1$ when $P_1=1 \ \text{atm}$ and $V_1= 1 \ \text{L}$ and then compute $T_2$ at $P_2=3 \ \text{atm}$, $V_2=3 \ \text{L}$, assuming a proportionality constant of $c=1$. Now I get the two temperatures $T_1=273 \ \text{K}$ and $T_2=2457 \ \text{K}$. So the difference between these temperatures is the added energy, how do I get it in Joules and is this method correct?

c) I feel like this question needs to be answered in order to be used in b), because in b), the quantity $Q$ is what I need, right?

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  • $\begingroup$ You can't get temperature from the information provided--you'd need to know how much helium is in the cylinder to do that. But you can get the right answer to (b) by the method you suggested; you could assume any value for $T_1$ and get the same result. $\endgroup$ – Ben51 Sep 10 '19 at 15:18
  • $\begingroup$ But we already know that we have 1 liter of Helium? $\endgroup$ – Parseval Sep 10 '19 at 15:22
  • $\begingroup$ That is the volume. But you could have one liter at a temperature of 100K or 1000K and still have a pressure of 1 atm. Just have to change the amount of gas (number of helium atoms). $\endgroup$ – Ben51 Sep 10 '19 at 15:24
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    $\begingroup$ Concerning the initial temperature, is it possible that the question was intended to mean that the cylinder starts out at room temperature but forgot to mention that? $\endgroup$ – user93237 Sep 10 '19 at 16:52
  • $\begingroup$ @SamuelWeir - Nope. I got it from this book, chek pdf-page 32. Question 1.31. islamabad-institute.weebly.com/uploads/1/3/1/0/13101414/… $\endgroup$ – Parseval Sep 10 '19 at 17:51
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You analyzed it correctly, and determined the final temperature correctly. Now just use $\Delta U=nC_v\Delta T$. You calculate the number of moles n from the initial condition.

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  • $\begingroup$ You can’t get temperature from information given. $\endgroup$ – Ben51 Sep 10 '19 at 17:12
  • $\begingroup$ This actually gives me the correct answer! But doesn't the formula you provided mean the process is Isobaric? Clearly, according to the problem the pressure varies. How can I explain this? $\endgroup$ – Parseval Sep 10 '19 at 17:55
  • $\begingroup$ @Ben51 - Since we are only interested in $\Delta T$ we can assume whatever first initial conditions. So it does work. $\endgroup$ – Parseval Sep 10 '19 at 17:59
  • $\begingroup$ Chet, do you agree with Parseval that you can just assume any initial temperature? Even if that can give you the internal energy change, it doesn't tell me what W or Q are since the process is not, in my opinion, clearly stated. It says pressure "rises" in proportion with volume. That implies a linear path- how do we get that for an ideal gas? To me, we don't know the path, we don't know the mass/moles, we only know the end points. What am I missing here? How do we get work and heat? $\endgroup$ – Bob D Sep 10 '19 at 18:07
  • $\begingroup$ @Parseval Chet's equation applies to any ideal gas, any process. $\endgroup$ – Bob D Sep 10 '19 at 18:08
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$W$ is work done on the system. $W=$-405 J

Since the gas expands, work is done by the system, not on the system. In order for the sign to be negative, you need to be using the chemist's version of the first law, namely

$$\Delta U=Q+W$$

and not the engineering/physics version of the first law, namely

$$\Delta U=Q-W$$

I noticed in the book you are using (pdf 28, pg 18) it is using the chemist version. If you wish to be consistent with the book, and get the correct answer for $Q$ now that @Chet Miller has shown you how to get $\Delta U$, revise your post to use $\Delta U=Q+W$,

Hope this helps.

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a) Calculate the work done by on the gas during this process, assuming that there are no other types of work being done.

You have already been told in connection with (a) that pressure cannot rise in direct proportion to volume in the case of an ideal gas. Pressure varies inversely with volume at constant temperature.

There are an infinite number of paths connecting the initial and final state. You cannot determine the work done unless you specify the path or have enough information to determine $\Delta U$ and $Q$ so you can determine $W$ by the first law.

b) Calculate the change in the helium's energy content during this process.

The change in internal energy of an ideal gas depends only on the temperature change according to

$$\Delta U=mC_{v}\Delta T$$

The only thing you can determine, from the ideal gas law given the initial and final states is the ratio of the final to initial temperature.

$$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$$

Since $P_2$=3$P_1$ and $V_2$=3$V_1$,

$$\frac{T_2}{T_1}=9$$

To get the actual temperatures you need either to know either the mass or the number of moles of helium

c) Calculate the amount of heat added to or removed from helium during this process

Since you have no legitimate information on the path from 1 to 2 and no information from which to compute the change the heat and work transfers. You may assume an initial temperature to the change in internal energy, but per the first law, $\Delta U=Q-W$, it doesn't tell you which part of the change is heat and which part work. You need more information.

Hope this helps.

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  • $\begingroup$ I think you are misreading the question. It is specified that the expansion is done in such a way that pressure rises in proportion with volume. This will require that heat is added during the process such that temperature rises quadratically with volume. There are unambiguous answers to these questions. $\endgroup$ – Ben51 Sep 10 '19 at 15:34
  • $\begingroup$ @Ben51 I am not misreading the question. The questions states the "pressure rises in proportion with volume". This definitely implies a linear path from 1 to 2 which is not possible for an ideal gas. Look at my answer to the OPs original post. It is a different matter to say that the ratio of pressure to volume is the same at the final state as the original state. That simply defines the equilibrium states, but it does not define the process. You say the "temperature rises quadratically with volume". $\endgroup$ – Bob D Sep 10 '19 at 16:07
  • $\begingroup$ That is not possible since the ideal gas equation clearly shows that, given the initial and final pressure and volumes, the final temperature has to be 3 times the initial temperature. Since the final volume is three times the initial volume, the ratio of temperature to volume is one. It is not a quadratic relationship. $\endgroup$ – Bob D Sep 10 '19 at 16:07
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    $\begingroup$ @BobD - I think that you made an arithmetic mistake between the second and third equations that you wrote down. The initial and final temperatures should differ by a factor of 9, not 3. $\endgroup$ – user93237 Sep 10 '19 at 17:07
  • $\begingroup$ @Samuel Weir Oops you’re right. Thanks $\endgroup$ – Bob D Sep 10 '19 at 17:43

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