# Volume ratio in an adiabatic gas expansion

• Granger
In summary, a perfect monoatomic gas at pressure $p_i$ 1.2 atm and temperature $T_i$ 300K undergoes an adiabatic expansion after a mass of 3.13kg is removed from a cylinder with a volume $V_i=1L$ and a piston with a mass of 1kg at a height of 50 cm. The final pressure is $p_f=1.04837$ and the final volume is $V_f$. The value of $Vi/Vf$ can be found using the equation for adiabatic volume changes and the fact that the gas is monatomic.
Granger

## 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?

Granger said:
Can someone give me a hint?

TSny said:

Yes I know that means the variation of internal energy is just the work done but how does that help...

@TSny if you want to be vague and leave without further explanations, please leave.

Granger said:
@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.

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.

Granger said:
Relevant equations
You should know an equation relating to adiabatic volume changes.

Granger said:
@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 realize 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 .

Last edited:
Granger and haruspex
@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.

Last edited:
Chestermiller
Granger said:
@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

haruspex said:
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.

Granger

## 1. What is the definition of volume ratio in an adiabatic gas expansion?

The volume ratio in an adiabatic gas expansion refers to the change in volume of a gas as it expands adiabatically, meaning without any heat exchange with the surroundings. It is typically represented by the symbol "V".

## 2. How is volume ratio related to temperature and pressure in an adiabatic gas expansion?

According to the ideal gas law, the volume of a gas is inversely proportional to its pressure, and directly proportional to its temperature. In an adiabatic gas expansion, the temperature of the gas decreases as it expands, causing the volume to increase. This results in a decrease in pressure, maintaining the inverse relationship between volume and pressure.

## 3. What is the formula for calculating volume ratio in an adiabatic gas expansion?

The formula for volume ratio in an adiabatic gas expansion is V2/V1 = (T2/T1)^γ, where V2 and V1 are the final and initial volumes, T2 and T1 are the final and initial temperatures, and γ is the adiabatic index or ratio of specific heats for the gas being expanded.

## 4. How does the adiabatic index or ratio of specific heats affect the volume ratio in an adiabatic gas expansion?

The adiabatic index or ratio of specific heats, denoted as γ, is a characteristic of a gas that determines its behavior during adiabatic processes. A gas with a higher value of γ will experience a larger change in temperature and pressure during an adiabatic expansion, resulting in a larger volume ratio.

## 5. Can the volume ratio in an adiabatic gas expansion ever be greater than 1?

Yes, the volume ratio in an adiabatic gas expansion can be greater than 1. This occurs when the initial temperature and pressure of the gas are high enough that the final temperature and pressure after expansion are still greater than atmospheric conditions. This is known as a superadiabatic expansion.

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