Derive ##\dfrac{\rho_1}{\rho_2}=\dfrac{T_2}{T_1}##

AI Thread Summary
The discussion revolves around deriving the relationship between the densities and temperatures of two gases, specifically in the context of a hot-air balloon. Participants express confusion about the definitions of the gases involved and their interactions, particularly regarding pressure and volume constraints. The fixed volume of the balloon is emphasized, with clarification that the balloon can still adjust the amount of gas inside as temperature changes. The conversation highlights the importance of understanding how temperature affects pressure in a closed system, especially in relation to buoyancy. Ultimately, the aim is to clarify the principles governing the behavior of gases in the balloon scenario.
Istiak
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Homework Statement
Derive ##\dfrac{\rho_1}{\rho_2}=\dfrac{T_2}{T_1}##
Relevant Equations
PV=nRT
I just found here(https://byjus.com/physics/relation-between-density-and-temperature/#:~:text=Density and Temperature Relationship 1 When density increases,,reduces. 4 When the temperature decrease, density increases.) that P=##rho##RT. So they just took ##\rho=\frac{n}{V}##.

##\dfrac{P_1}{P_2}=\dfrac{\rho_1 RT_1}{\rho_2 RT_2}##
##T_2=\frac{\rho_1T_1P_2}{\rho_2P_1}##
I got pressure while I want to ignore it. No, I don't want to choose ##P_n##=1
 
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What are 1 and 2 defined in the problem ?
 
anuttarasammyak said:
What are 1 and 2 defined in the problem ?
first object and 2nd object. e.g. ##\rho_1## density of first object.
 
Thanks. How do these objects are connected or coexist, the problem says ?
 
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anuttarasammyak said:
Thanks. How do these objects are connected or coexist, the problem says ?
##\rho_1=n\rho_2##.
 
n? Number of particles? Which particles of objects1 or 2 or ... ?
 
anuttarasammyak said:
n? Number of particles? Which particles of objects1 or 2 or ... ?
No. ##n## is just an integer.
 
I am still puzzled. There are gas #1 and gas #2. Are they independent ? mixed in a same vessel ? pushing a piston from right and left ? or ... I do not get the situation.
 
The problem said that, "density of first object is ##n## times of second object". Hope it's clearer more now
 
  • #10
Consider a hot-air balloon with fixed volume VB = 1.1 m3. The mass of the balloonenvelope, whose volume is to be neglected in comparison to VB, is mH = 0.187 kg. The balloon shall be started, where the external air temperature is ϑ1 = 20 oC and the normal external air pressure is po = 1.013 ⋅ 105 Pa. Under these conditions the density of air is ρ1 = 1.2 kg/m3.

What temperature ϑ2 must the warmed air inside the balloon have to make the balloon just float?
<br/>
sh*t I mixed up two questions. That's why you were getting confused, Sorry for that.
 
  • #11
So gas #1 is the Air and gas #2 seems to be gas in the balloon, right ? You could find the relation between ##P_1## and ##P_2##.
 
  • #12
anuttarasammyak said:
So gas #1 is the Air and gas #2 seems to be gas in the balloon, right ? You could find the relation between ##P_1## and ##P_2##.
Are you talking about ##(\dfrac{V_1}{V_2})^\gamma=\dfrac{P_2}{P_1}##?
 
  • #13
When gas pressures inside and outside balloon are different, the balloon inflates or shrinks for the pressures to be ...
 
  • #14
anuttarasammyak said:
When gas pressures inside and outside balloon are different, the balloon inflates or shrinks for the pressures to be ...
I think that depends. Like, if outside pressure is higher than it shrinks and inflates for less pressure outside
 
  • #15
Right. So how do you estimate the pressures inside and outside in the end ?
 
  • #16
anuttarasammyak said:
Right. So how do you estimate the pressures inside and outside in the end ?
don't know.

But I may try, at the end I will consider their temperature is same. So I will start with ##PV=nRT##
##\frac{P_1V_1}{P_2V_2}=1##? Considered for a single mole ##n=1##.
 
  • #17
anuttarasammyak said:
When gas pressures inside and outside balloon are different, the balloon inflates or shrinks for the pressures to be ...
Yet the statement of the problem says that the volume is "fixed".
Istiakshovon said:
Consider a hot-air balloon with fixed volume VB = 1.1 m3.
I understand that the balloon will not float unless its volume increases. I do not understand how to interpret "fixed". Maybe the 1.1 m3 is an upper limit?

Edit: The volume is constant. As the temperature of the air inside changes, air will leave or enter the balloon.
 
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  • #18
kuruman said:
Yet the statement of the problem says that the volume is "fixed".
Thanks for reminding. So 1.1m^3 gas inside the balloon should be lighter than 1.1 m^3 air around with 0.187 kg for just float.
 
  • #19
Istiakshovon said:
Homework Statement:: Derive ##\dfrac{\rho_1}{\rho_2}=\dfrac{T_2}{T_1}##
Relevant Equations:: PV=nRT
In order to get a hint how about thinking other way around, i.e.,
When ##\dfrac{\rho_1}{\rho_2}=\dfrac{T_2}{T_1}##, get the relation of ##P_1## and ##P_2## ?
 
  • #20
##\rho=\frac{P}{RT}##
##\dfrac{\rho_1}{\rho_2}=\frac{T_2}{T_1}##
##\dfrac{P_1RT_2}{RT_1P_2}=\frac{T_2}{T_1}##
##\dfrac{P_1}{P_2}=1##
?
 
  • #21
Istiakshovon said:
##\rho=\frac{P}{RT}##
##\dfrac{\rho_1}{\rho_2}=\frac{T_2}{T_1}##
##\dfrac{P_1RT_2}{RT_1P_2}=\frac{T_2}{T_1}##
##\dfrac{P_1}{P_2}=1##
?
Why the question mark? That says that the pressure inside and outside the balloon is the same. Doesn't make sense? The volume enclosed by the balloon envelope is fixed but not the amount of air inside. Have you ever seen a hot-air balloon? The envelope has an opening at the bottom and a heater directly underneath the opening. How does that work?
 
  • #22
kuruman said:
Why the question mark?
The equation on #20 were saying pressures are same. But in #16 I got that pressure times volume are the same. If we consider pressure to be same then Volume should be same. That's what was confusing me that moment.

kuruman said:
How does that work?
The fire increases the pressure of the balloon if we consider volume to be constant. From ##PV=nRT##, pressure increases as the temperature increases.
 
  • #23
Istiakshovon said:
The equation on #20 were saying pressures are same. But in #16 I got that pressure times volume are the same. If we consider pressure to be same then Volume should be same. That's what was confusing me that moment.The fire increases the pressure of the balloon if we consider volume to be constant. From ##PV=nRT##, pressure increases as the temperature increases.
You don't understand how a hot air balloon works and you didn't read post #21 carefully enough. Yes, ##pV=nRT##. The volume ##V## is fixed and the balloon is open at the bottom. Please answer the following questions in sequence:
1. How does the pressure inside the balloon differ from the pressure outside? Remember the balloon is open to the atmosphere.
2. What does ##pV=nRT## predict will happen when you raise the temperature inside from ##T_1## to ##T_2## keeping the volume constant?
3. Is your answer in question 2 consistent with your answer in question 1?
 
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