Relationship between pressure and volume

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Homework Help Overview

The discussion revolves around the relationship between pressure and volume of an ideal gas under isothermal conditions, specifically comparing two different masses of the same gas at a given temperature.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the Ideal Gas Law and question the relationship between temperature, pressure, and mass. There are attempts to clarify whether the temperatures of the gases are indeed equal and how this affects the comparison of masses.

Discussion Status

There is an ongoing exploration of the assumptions regarding temperature and mass, with some participants suggesting that the isotherms represent different masses at the same temperature. Others are questioning the validity of these assumptions and the implications of the Ideal Gas Law.

Contextual Notes

Participants note the importance of understanding isotherms and the conditions under which the Ideal Gas Law applies, particularly regarding the fixed nature of temperature and the implications for mass comparison.

Amith2006
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Sir,
Two different isothermal curves representing the relationship between pressure and volume at a given temperature of the same ideal gas are shown for masses M1 and M2 of the gas respectively in the following figure, then which is greater M1 or M2?
 

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a) Please show your work first.

b) I can't see the attachment, but here's a hint : what does the Ideal Gas Law state? How does this bear on your problem?
 
Sir,
I solved it the following way:
From the figure T1>T2. But according to ideal gas law, Temperature is inversely proportional to mass of gas. Therefore M1<M2.Is it right?
 
But the question states that the ideal gas is at a given temperature. Doesn't this mean that T1=T2?
 
The ideal gas equation says
[tex]PV = nRT[/tex]
The number of moles of the gas is given by (m = mass of gas, M = molar mass)
[tex]n = \frac{m}{M}[/tex]
So
[tex]PV = \frac{m}{M}RT[/tex]
Solving for m, the mass of the gas
[tex]m = \frac{PV}{RT}M[/tex]

If you draw a vertical line (parallel to the P-axis) it cuts the two curves at pressures [itex]P_{2}[/itex] and [itex]P_{1}[/itex]. Clearly [itex]P_{2} > P_{1}[/itex]. Also [itex]T_{2} > T_{1}[/itex] (can you see why?). I think I have said more than enough.

Amit, it is not correct to say that the temperature is inversely or directly proportional to the mass of the gas unless pressure and volume are constant. As you can see here, pressure and temperature are different for the two gases.
 
pizzasky said:
But the question states that the ideal gas is at a given temperature. Doesn't this mean that T1=T2?

Do you know what an isotherm is?
 
I would agree with Pizzasky's interpretation of the problem. The two graphs are the isotherms for two different quantities of the same gas for the same temperature. Since PV (the area of the rectangle with corners on the origin and on the curve) is greater for M2, this means that n2RT > n1RT so M2>M1.

AM
 
If you say that the temperatures are equal, then you and Pizzasky are correct. It seems that the wording suggests so. Fine.
 
Maverick,

I agree with AM and Pizza. If the number of moles of the gas is unknown, you can't say that the outer isotherm corresponds to a higher temperature. An isotherm that is farther from the origin can also be generated at the same temperature by increasing the number of moles in the sample.

In this case, if you allow the number of moles (through the total mass) and the temperature to float, you can't make any meaningful comparison. One of the two must be fixed, and that would be the temperature.

Edit : I wrote this before I saw your last post.
 

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