Maximum temperature reached by gas in expansion

[timetraveller123]

Homework Statement

1 mole of ideal gas with internal energy U= 3/2 RT , expands from initial volume V_i = 1/10 V_o following the equation p=(− p_o / V_o ) V +p_o
.
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=(− p_o / V_o ) V +p_o
pV=(− p_o / V_o ) V² +p_oV
(pV/nr)=( (− p_o / V_o ) V² +p_oV)/nr = T
taking derivative of T with respect to v gets
dT/dx= ((− p_o / V_o ) 2V +p_o)/nr
setting derivative to 0 yields
V = V_o/2
plugging back into equation for temperature gets
maximum temperature = p_oV₀/4nr
[/B]
this is what i got is the right method or is there a way to get a numerical anser

 

[kuruman]

The best you can do is express the maximum temperature as a number times the initial temperature.

 

[timetraveller123]

ok then how do i do part b

 

[kuruman]

ok then how do i do part b

Use the first law.

 

[timetraveller123]

you mean llike Δu + w =q

 

[timetraveller123]

as Δu is easy but w = ∫(− p_o / V_o ) V +p_o 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

 

[Chestermiller]

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?

 

[timetraveller123]

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 p_o and v_o are not the initial states they are just some constants
initial volume = 1/10 v_o
initial pressure can be calculated using the above equation
thus maximum temperature reached is 100/36 T_i

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?

 

[Chestermiller]

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

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.

 

[timetraveller123]

∫(− po / Vo ) V +po dv = w.
[ (− p₀ / V₀ ) V²/2 + p₀V ]_vi^v_f
while v_i is known what do i plug in for v_f 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!

 

[Chestermiller]

∫(− po / Vo ) V +po dv = w.
[ (− p₀ / V₀ ) V²/2 + p₀V ]_vi^v_f
while v_i is known what do i plug in for v_f 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.

 

[timetraveller123]

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

 

[Chestermiller]

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?

 

[timetraveller123]

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

 

[Chestermiller]

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?

 

[timetraveller123]

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

 

[Chestermiller]

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.

 

[timetraveller123]

ooh okay how did you arrive at that thanks anyways

 

[Chestermiller]

ooh okay how did you arrive at that thanks anyways

It's the only thing that makes sense to me.

 

[kuruman]

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.

 

[timetraveller123]

so basically i just plug in v for maximum temperature

 

[kuruman]

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.

 

[timetraveller123]

ok sure i will do that give me some time i am very busy

 

[kuruman]

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.

 

[Chestermiller]

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.

 

[timetraveller123]

(pV/nr)=( (− p_o / V_o ) V² +p_oV)/nr = T_final
for T_initial i pugged in V = 1/10 V₁₀ as given
(pV/nr)=( (− p_o ) V_o/100 +p_oV_o/10)/nr = T_inital
hence Δu becomes 3/2nr(T_final - T_initial)
Δu = 3/2(( (− p_o / V_o ) V² +p_oV)− p_o V_o/100 +p_oV_o/10)

w = ∫(− po / Vo ) V +po dv

Q = Δu +w

setting derivative of the expression to zero we get

0 = 3/2(-2p_oV/V_o = p_o) - p_oV/v_o + P_o

solving for v i get 5/8V_o = 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

 

[Chestermiller]

(pV/nr)=( (− p_o / V_o ) V² +p_oV)/nr = T_final
for T_initial i pugged in V = 1/10 V₁₀ as given
(pV/nr)=( (− p_o ) V_o/100 +p_oV_o/10)/nr = T_inital
hence Δu becomes 3/2nr(T_final - T_initial)
Δu = 3/2(( (− p_o / V_o ) V² +p_oV)− p_o V_o/100 +p_oV_o/10)

w = ∫(− po / Vo ) V +po dv

Q = Δu +w

setting derivative of the expression to zero we get

0 = 3/2(-2p_oV/V_o = p_o) - p_oV/v_o + P_o

solving for v i get 5/8V_o = 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.

 

[timetraveller123]

thanks so much for the help