Confused about the ideal gas law

In summary, the conversation discusses the struggles of understanding physics compared to math and a specific problem involving finding the density of air in two different states. The incorrect assumption of density being a constant property is mentioned and the correct equation for density of an ideal gas is provided. The conversation ends with a clarification that density of air is not always 1.22kg/m3 and can vary depending on the conditions.
  • #1
LT72884
323
48
Ok, i am struggling to figure something out. I don't know why math is so much easier than physics haha.

ok, here is my struggle. I have two states, state 1 and 2, which i will call just 1 and 2.

1:
T=298kelvin
V=0.025m(cubed)
P=310Kpa
Mass1=Mass2
R=0.2870

2:
T=323kelvin
V=0.025m(cubed)
P=??
Mass2 = Mass1
R=0.2870

This is how i solved for P2 and it was WRONG!

P2V2=mR2T2 where m is mass

mass=V(rho) = 0.025(1.22) = 0.0305kg of air in the closed system

P2= mRT divided by volume. I get 113kpa. Book says 336kpa.

IF I USE P1V1/T1=P2V2/T2, i get the right answer. BUT it should have worked the way i did solving for mass because mass IS volume multiplied by density and it doesn't change..

Thanks
 
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  • #2
LT72884 said:
mass=V(rho) = 0.025(1.22)
Under what conditions of pressure and temperature is ρair = 1.22 kg/m3 ?
It is apparent that an incorrect assumption is being made here.

Find the actual density from the information given for State 1.
 
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  • #3
256bits said:
Under what conditions of pressure and temperature is ρair = 1.22 kg/m3 ?
It is apparent that an incorrect assumption is being made here.

Find the actual density from the information given for State 1.
to answer your first question, i thought density never changes for a substance. that is why it is called an intensive property? its ALWAYS 1000kg/m3 or 1.22kg/m3, etc etc. because if it were to change, then all my problems i did for finding pressure in a multifluid manometer, would be wrong because we assumed rho of water to be 1000kg/m3, air to be 1.22kg/m3 even when Patm was lower or higher than 1atm

How would i find density in state one if all i have is volume? m=PV/RT so rho = PV/RT divided by Volume? mass = 0/0906kg divided by 0.025 = 3.62kg/m3.. does not make sense to me. I have always been told that air is 1.22kg/m3 no mater what. that's how we did our multifluid manometers and all of my water pressure problems that included a Patm different than 1atm

thanks for helping me understand this.

EDIT:

I should add. I am not saying you are wrong at all. Just saying it makes no sense to me haha.
 
  • #4
LT72884 said:
Just saying it makes no sense to me
If you have a bicycle pump, close off the valve end with your finger, and press down on the handle.
The mass in the cylinder doesn't change, but the volume does.
The density of the air within the cylinder does what?

another example,
The atmosphere ( of the Earth ) as one goes farther up from that at the surface of the Earth to the vacuum of space.
Density of the air does what as one increases in altitude?
LT72884 said:
have always been told that air is 1.22kg/m3 no mater what
"No matter what" is a very strong statement to make.
 
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  • #5
LT72884 said:
i thought density never changes for a substance. that is why it is called an intensive property
That's not the meaning of an intensive property.

Temperature is an intensive property, but we all know that the day to day outside temperature changes.
A review of this subject t- intensive and extensive properties - might help.
 
  • #6
256bits said:
If you have a bicycle pump, close off the valve end with your finger, and press down on the handle.
The mass in the cylinder doesn't change, but the volume does.
The density of the air within the cylinder does what?

another example,
The atmosphere ( of the Earth ) as one goes farther up from that at the surface of the Earth to the vacuum of space.
Density of the air does what as one increases in altitude?
"No matter what" is a very strong statement to make.

yeah, the volume of air changes(decreases), and not the mass since none escaped. so density would be different then.

yes, "no matter what" is a strong statement and that is why i am confused haha. because all the basic physics stuff i took is now being challenged in thermo haha
 
  • #7
LT72884 said:
to answer your first question, i thought density never changes for a substance. that is why it is called an intensive property?
I think you might be confusing “intensive property” with “intrinsic property”, maybe?
 
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  • #8
256bits said:
That's not the meaning of an intensive property.

Temperature is an intensive property, but we all know that the day to day outside temperature changes.
A review of this subject t- intensive and extensive properties - might help.
true, but if i take a room at 72F and put a wall in the center to divide it in half, the temp on both sides would be the same 72F. the volume of air would be diffeernet than the original value, but the density would be the same and so would the specific volume as well as the mass.
 
  • #9
Since you think math is so much easier than physics, here is the math equation for the density of an ideal gas:
$$\rho=\frac{PM}{RT}$$where M is the molar mass of the gas.
 
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  • #10
Chestermiller said:
Since you think math is so much easier than physics, here is the math equation for the density of an ideal gas:
$$\rho=\frac{PM}{RT}$$where M is the molar mass of the gas.
Thank you:) this is one of yhe equations i have in my notes and book. I guess where my confusion was is the fact that density of air is not always 1.22kg/m3 haha
 
  • #11
Unless they want you to solve for the mass of the air, then you can get by without solving for it.
With ## P*V = m*R*T ## and you have established that mas is constant, so ##m*R## is also a constant.
 
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  • #12
scottdave said:
Unless they want you to solve for the mass of the air, then you can get by without solving for it.
With ## P*V = m*R*T ## and you have established that mas is constant, so ##m*R## is also a constant.
Correct. I can use pv/t for my calculations. I was just stuck on the fact that air is always 1.22kg/m3 but it aint haha
 
  • #13
If you need the mass, you could just solve for it from the equation P1V1 = m r T1, where you are given everything except m. Also, it is a good idea to include units in all of your calculations, to help keep yourself straight. This one is not too complex, but you may run into ones in the future which get pretty complex
 
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  • #14
The density of a liquid such as water doesn't change by "much" with temperature compared to gasses. But they do change. Even solids expand and contract (same mass but different volume) some with temperature. Any density you see for a material should list the conditions which this value is.
 
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  • #15
scottdave said:
If you need the mass, you could just solve for it from the equation P1V1 = m r T1, where you are given everything except m. Also, it is a good idea to include units in all of your calculations, to help keep yourself straight. This one is not too complex, but you may run into ones in the future which get pretty complex
I tried that and ended up with a number not close at all to what was needed haha
 
  • #16
scottdave said:
The density of a liquid such as water doesn't change by "much" with temperature compared to gasses. But they do change. Even solids expand and contract (same mass but different volume) some with temperature. Any density you see for a material should list the conditions which this value is.
This is all good to know haha. Breaking the old indoctrinated info with new info haha
 
  • #17
You might try this. p(1)V(1) = n(1)RT(1) and p(2)V(2) = n(2)RT(2) and V(1) = V(2) and n(1) = n(2). Divide eqn, 2 by eqn. 1 and solve for T(2).
 

1. What is the ideal gas law and why is it important in science?

The ideal gas law is a mathematical equation that describes the behavior of an ideal gas at a given temperature, pressure, and volume. It is important in science because it allows us to predict the behavior of gases in various conditions and is used in many scientific fields, including chemistry, physics, and engineering.

2. What are the variables in the ideal gas law and how are they related?

The variables in the ideal gas law are pressure (P), volume (V), temperature (T), and the number of moles of gas (n). They are related through the equation PV = nRT, where R is the gas constant. This equation shows that as one variable changes, the others will also change in a predictable way.

3. Can the ideal gas law be applied to real gases?

The ideal gas law is an approximation and is most accurate for gases at low pressures and high temperatures. Real gases deviate from ideal behavior at high pressures and low temperatures, so corrections must be made to the ideal gas law to accurately describe their behavior.

4. How is the ideal gas law used in practical applications?

The ideal gas law is used in many practical applications, such as in the design of gas storage tanks and in the production of industrial gases. It is also used in weather forecasting to predict changes in air pressure and volume. Additionally, it is used in the study of gas behavior in chemical reactions and in the development of new materials.

5. What are some common misconceptions about the ideal gas law?

One common misconception about the ideal gas law is that it only applies to gases at standard temperature and pressure (STP). In reality, the ideal gas law can be applied to gases at any temperature and pressure, as long as they are in a gaseous state. Another misconception is that real gases behave exactly like ideal gases, when in fact, they deviate from ideal behavior under certain conditions.

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