Volume ratio in an adiabatic gas expansion

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Homework Statement


Consider a perfect monoatomic gas at pressure $p_i$ 1.2 atm and temperature $T_i$ 300K, that is in equilibrium inside a cylinder having a volume $V_i=1L$ and which piston has a mass of 1kg and is at an height of 50 cm. Admit that a mass M=3.13kg is over the piston. When that mass is removed, the gas suffers an adiabatic expansion until a final pressure pf=1.04837, a final temperature Tf and a final volume Vf.

What is the value of Vi/Vf?

Homework Equations


3. The Attempt at a Solution [/B]
So my immediate thought was to use the equation of real gases
$$pV=nRT$$
´
We will have then
$$\frac{p_iV_i}{T_i}=\frac{p_fV_f}{T_f}$$

But I also don't know Tf so I'm stuck. Can someone give me a hint?
 

Answers and Replies

  • #2
TSny
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Can someone give me a hint?
The expansion is adiabatic.
 
  • #3
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The expansion is adiabatic.

Yes I know that means the variation of internal energy is just the work done but how does that help...
 
  • #4
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@TSny if you want to be vague and leave without further explanations, please leave.
 
  • #5
@TSny if you want to be vague and leave without further explanations, please leave.
Forum rules prevent giving very explicit answers. In fact I sometimes think I'm too explicit. :smile::smile:

There is another important fact in the problem: gas is monatomic.
Now with this information, what I would do would be to search in Google how the adiabatic processes behave.
 
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@TSny if you want to be vague and leave without further explanations, please leave.

@Granger ,

You need to have some patience and little more decency if you wish to interact on this forum .

You are wrong on several accounts .

1) You said , TSny is vague . I don't think so . You asked for a hint . He gave you a hint . You might not have found that hint helpful . In that case you could have requested him to elaborate his hint . He would have surely given you a more useful hint .

2) You presumed that he left you without further explanation .Do you realise your 2nd comment after TSny's reply came within 11 minutes of your 1st comment ? He might not have seen your 1st comment at all . Or even if had seen your reply he might have been in the process of replying you .It takes time to think and type in the response . Why be so impatient ?

Members here are not paid . It's their choice to help . You can't force somebody .

Even if you required an additional hint , there is a polite way to ask .

3) Asking TSny to leave , now that is so rude . Have some decency .This is not the way to react on a forum .

Presuming you have not interacted with TSny before, let me tell you he is a top class contributer on PF , surely one of the best I have seen .

So next time you write something on this forum , please show patience and decency .

Hope you will take this message in the right spirit .
 
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@TSny I want to apologize for my reaction on the other day. I was very stressed because I've been working for hours not figuring out how to solve the problem. Not that that can justify my rudeness. So my sincere apologies. I eventually figured out what yo meant with the hint that it was an adiabatic process (Q=0). I was not figuring out the expression for U in the case of an ideal monoatomic gas so I wasn't able to relate it with the work expression.
 
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  • #9
haruspex
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@Tsnys I want to apologize for my reaction on the other day. I was very stressed because I've been working for hours not figuring out how to solve the problem. Not that that can justify my rudeness. So my sincere apologies. I eventually figured out what yo meant with the hint that it was an adiabatic process (Q=0). I was not figuring out the expression for U in the case of an ideal monoatomic gas so I wasn't able to relate it with the work expression.
@TSny
 
  • #10
TSny
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No problem. Apology accepted. After posting my first reply, I was distracted by some personal matters. So I was not able to keep up my end of the conversation.Sorry for that. Glad you were able to solve the exercise.
 

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