Maximum temperature reached by gas in expansion

In summary: The problem is worded poorly. They want you to calculate the heat when the temperature reaches its maximum.I would plug in the volume that gives the maximum temperature.
  • #1
timetraveller123
621
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Homework Statement


1 mole of ideal gas with internal energy U= 3/2 RT , expands from initial volume Vi = 1/10 Vo following the equation p=(− po / Vo ) V +po
.
Find
(a) the highest temperature reached by the gas during the expansion and
(b) the maximum amount of heat taken in by the gas.

Homework Equations


pv = nrt

The Attempt at a Solution


p=(− po / Vo ) V +po
pV=(− po / Vo ) V2 +poV
(pV/nr)=( (− po / Vo ) V2 +poV)/nr = T
taking derivative of T with respect to v gets
dT/dx= ((− po / Vo ) 2V +po)/nr
setting derivative to 0 yields
V = Vo/2
plugging back into equation for temperature gets
maximum temperature = poV0/4nr
[/B]
this is what i got is the right method or is there a way to get a numerical anser
 
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  • #2
The best you can do is express the maximum temperature as a number times the initial temperature.
 
  • #3
ok then how do i do part b
 
  • #4
vishnu 73 said:
ok then how do i do part b
Use the first law.
 
  • #5
you mean llike Δu + w =q
 
  • #6
as Δu is easy but w = ∫(− po / Vo ) V +po dv
so solving the integral do i plug in v as the v obtained for maximum temperature or what?
is most heat added into reach the highest temperature in this case
 
Last edited:
  • #7
You calculated that the highest temperature is 1/4 the initial temperature. How can that be if the initial temperature is one of the states that it passes through? Regarding application of the first law, why do't you just run the calculation and see what you get?
 
  • #8
Chestermiller said:
You calculated that the highest temperature is 1/4 the initial temperature. How can that be if the initial temperature is one of the states that it passes through? Regarding application of the first law, why do't you just run the calculation and see what you get?

no in this question po and vo are not the initial states they are just some constants
initial volume = 1/10 vo
initial pressure can be calculated using the above equation
thus maximum temperature reached is 100/36 Ti

now my my question is
is maximum heat added to bring the system to state with highest temperature?
if that is so then the problem is very easy hope you understand my question?
 
  • #9
vishnu 73 said:
no in this question po and vo are not the initial states they are just some constants
initial volume = 1/10 vo
initial pressure can be calculated using the above equation
thus maximum temperature reached is 100/36 Ti
Oops. Sorry. My mistake.
now my my question is
is maximum heat added to bring the system to state with highest temperature?
if that is so then the problem is very easy hope you understand my question?
The question is kind of ambiguous. I would just solve the problem as a function of V and see how it plays out. Let the math do the work for you.
 
  • #10
(− po / Vo ) V +po dv = w.
[ (− p0 / V0 ) V2/2 + p0V ]vivf

while vi is known what do i plug in for vf is it the volume that gives maximum temperature or what or should i differentiate the above integral to find when the derivative of work is zero and plug in that v. thanks!
 
  • #11
vishnu 73 said:
(− po / Vo ) V +po dv = w.
[ (− p0 / V0 ) V2/2 + p0V ]vivf

while vi is known what do i plug in for vf is it the volume that gives maximum temperature or what or should i differentiate the above integral to find when the derivative of work is zero and plug in that v. thanks!
I would plug in the volume that gives the maximum temperature.
 
  • #12
but how can we be sure that yields maximum works and is the other method just as equally correct
and btw thanks for the fast replies
 
  • #13
vishnu 73 said:
but how can we be sure that yields maximum works and is the other method just as equally correct
and btw thanks for the fast replies
Who said anything about maximum work?
 
  • #14
no because q = Δu + w
so isn't maximum q added when both these quantities are maximum am i wrong please correct me if i am
 
  • #15
vishnu 73 said:
no because q = Δu + w
so isn't maximum q added when both these quantities are maximum am i wrong please correct me if i am
Isn't there also a change in internal energy?
 
  • #16
yeah change in internal energy is automatically maximum when the temperature is maximum but how are we supposed to know that is the case for work done by the gas too
 
  • #17
vishnu 73 said:
yeah change in internal energy is automatically maximum when the temperature is maximum but how are we supposed to know that is the case for work done by the gas too
The problem is worded poorly. They want you to calculate the heat when the temperature reaches its maximum.
 
  • #18
ooh okay how did you arrive at that thanks anyways
 
  • #19
vishnu 73 said:
ooh okay how did you arrive at that thanks anyways
It's the only thing that makes sense to me.
 
  • #20
What if you used the First Law to find an expression for the heat Q in terms of independent variable V and constants p0 and V0? Then you can find the value of V at which the heat that has entered the gas up to that point reaches a maximum before it starts going down, i.e. before heat starts being removed from the gas. The answer is a simple fraction of V0.
 
  • #21
so basically i just plug in v for maximum temperature
 
  • #22
vishnu 73 said:
so basically i just plug in v for maximum temperature
That's not it. First find an expression for Q in terms on V, then maximize that with respect to V. To do this, you need to find ΔU and W in terms of V.
 
  • #23
ok sure i will do that give me some time i am very busy
 
  • #24
vishnu 73 said:
ok sure i will do that give me some time i am very busy
Take all the time you need, this is a web-based help session, not an examination.
 
  • #25
I stand corrected. kuruman is right. I solved this problem, and found that the volume at which the temperature is maximum is different from the volume at which the heat added is maximum.
 
  • #26
(pV/nr)=( (− po / Vo ) V2 +poV)/nr = Tfinal
for Tinitial i pugged in V = 1/10 V10 as given
(pV/nr)=( (− po ) Vo/100 +poVo/10)/nr = Tinital
hence Δu becomes 3/2nr(Tfinal - Tinitial)
Δu = 3/2(( (− po / Vo ) V2 +poV)− po Vo/100 +poVo/10)

w = ∫(− po / Vo ) V +po dv

Q = Δu +w

setting derivative of the expression to zero we get

0 = 3/2(-2poV/Vo = po) - poV/vo + Po

solving for v i get 5/8Vo = v

am i right is this the volume for which maximum q is added after this i should be pretty much able to do myself

and sorry for the late reply
 
  • #27
vishnu 73 said:
(pV/nr)=( (− po / Vo ) V2 +poV)/nr = Tfinal
for Tinitial i pugged in V = 1/10 V10 as given
(pV/nr)=( (− po ) Vo/100 +poVo/10)/nr = Tinital
hence Δu becomes 3/2nr(Tfinal - Tinitial)
Δu = 3/2(( (− po / Vo ) V2 +poV)− po Vo/100 +poVo/10)

w = ∫(− po / Vo ) V +po dv

Q = Δu +w

setting derivative of the expression to zero we get

0 = 3/2(-2poV/Vo = po) - poV/vo + Po

solving for v i get 5/8Vo = v

am i right is this the volume for which maximum q is added after this i should be pretty much able to do myself

and sorry for the late reply
Yes, correct volume for max Q.
 
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Likes timetraveller123
  • #28
thanks so much for the help
 

1. What is the maximum temperature reached by gas during expansion?

The maximum temperature reached by gas during expansion is known as the adiabatic flame temperature. It is the highest temperature that can be achieved by a perfect combustion process without any heat loss to the surroundings.

2. How is the maximum temperature of gas in expansion calculated?

The maximum temperature of gas in expansion can be calculated using the adiabatic flame temperature equation, which takes into account the initial temperature, the specific heat ratio, and the initial and final pressures of the gas.

3. What factors affect the maximum temperature of gas in expansion?

The maximum temperature of gas in expansion is affected by the type of gas, its initial temperature and pressure, the specific heat ratio, and the conditions of the expansion process (adiabatic or isothermal).

4. Why is the maximum temperature of gas in expansion important?

The maximum temperature of gas in expansion is important in various industrial processes, such as combustion engines, gas turbines, and rocket engines. It helps in determining the efficiency and performance of these processes and can also impact the safety and stability of the system.

5. How can the maximum temperature of gas in expansion be controlled?

The maximum temperature of gas in expansion can be controlled by adjusting the initial conditions of the gas, such as its temperature and pressure, and by controlling the expansion process. For example, in an adiabatic expansion, adding a heat exchanger can help in reducing the temperature. In an isothermal expansion, controlling the rate of expansion can help in controlling the maximum temperature.

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