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

In summary: That's why it floats. The balloon is pushed up by the excess pressure of the air inside over the air outside. The envelope of the balloon is not even airtight so some air can leak out. The air inside is heated by the fire. As the air inside gets hotter, the air outside gets cooler. The heated air is less dense than the outside air. When the air inside is heated enough, the balloon floats. When the air cools enough the balloon sinks.In summary, the conversation discusses the relationship between density and temperature, and how it is related to pressure and volume in a hot-air balloon. The equation P=ρRT is used to calculate
  • #1
Istiak
158
12
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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  • #2
What are 1 and 2 defined in the problem ?
 
  • #3
anuttarasammyak said:
What are 1 and 2 defined in the problem ?
first object and 2nd object. e.g. ##\rho_1## density of first object.
 
  • #4
Thanks. How do these objects are connected or coexist, the problem says ?
 
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  • #5
anuttarasammyak said:
Thanks. How do these objects are connected or coexist, the problem says ?
##\rho_1=n\rho_2##.
 
  • #6
n? Number of particles? Which particles of objects1 or 2 or ... ?
 
  • #7
anuttarasammyak said:
n? Number of particles? Which particles of objects1 or 2 or ... ?
No. ##n## is just an integer.
 
  • #8
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.
 
  • #9
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?
 

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

What is the meaning of the symbols in the equation?

The symbol ##\rho## represents density, while ##T## represents temperature. The subscripts 1 and 2 indicate two different states or conditions.

What is the significance of the ratio of densities in this equation?

The ratio of densities in this equation is known as the density ratio. It is a fundamental concept in fluid mechanics and thermodynamics, and it describes the relationship between the density of a fluid at two different states or conditions.

How is this equation derived?

This equation is derived from the ideal gas law, which states that the product of pressure and volume is proportional to the product of the number of moles of gas and the temperature. By rearranging this equation and considering the relationship between density and mass, we arrive at the equation ##\dfrac{\rho_1}{\rho_2}=\dfrac{T_2}{T_1}##.

What are the applications of this equation?

This equation has many applications in various fields, including fluid mechanics, thermodynamics, and meteorology. It is used to study the behavior of gases in different conditions, such as in engines, pipelines, and weather systems.

Are there any limitations to this equation?

While this equation is useful in many situations, it assumes that the gas is ideal and that the temperature and pressure are constant. In reality, these conditions may not always be met, and the equation may not accurately describe the behavior of real gases.

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