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Pressure Cooker

by CurtisB
Tags: cooker, pressure
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CurtisB
#1
May10-07, 09:48 AM
P: 16
1. The problem statement, all variables and given/known data
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping.


1-If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F_120 on the lid when the air inside the cooker had been heated to 120 degrees C Assume that the temperature of the air outside the pressure cooker is 20degrees C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be p_a.

Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying pV=N*k_B*T.

Express the force in terms of given variables.

2- The pressure relief valve on the lid is now opened, allowing hot air to escape until the pressure inside the cooker becomes equal to the outside pressure p_a. The pot is then sealed again and removed from the stove. Assume that when the cooker is removed from the stove, the air inside it is still at 120 degrees C.

What is the magnitude of the net force F_20 on the lid when the air inside the cooker has cooled back down to 20 degrees C?
Express the magnitude of the net force in terms of given variables.

2. Relevant equations
delta_U = n*C_v*delta_T
pV = N*k_B*T
F=pA

3. The attempt at a solution
I am realy stuck with this one, at first I tried re-arranging the equation given in terms of V to equate final and initial but that didn't work, can someone please give me a push in the right direction.
Cheers.
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andrevdh
#2
May10-07, 10:06 AM
HW Helper
P: 1,449
Quote Quote by CurtisB View Post
1. The problem statement, all variables and given/known data
A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping.


1-If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F_120 on the lid when the air inside the cooker had been heated to 120 degrees C Assume that the temperature of the air outside the pressure cooker is 20degrees C (room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be p_a.

Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying pV=N*k_B*T.

Express the force in terms of given variables.

Since V,N and k_B (R) is unaltered by this process of heating the air up inside of the pot we have that

[tex]\frac{p}{T} = constant[/tex]

which is known as the Guy-Lussac law.
CurtisB
#3
May10-07, 09:39 PM
P: 16
Thanks, that really helped, I got the answer to part one to be 0.34Ap_a but now the second part is getting me. I work out the final pressure inside to be
(293.15p_a)/(393.15) so the force out = (293.15Ap_a)/(393.15), then the force in equals -Ap_a so the final answer should be (293.15Ap_a)/(393.15)-Ap_a, right, but apparently I have the signs mixed up, am I right in saying that net force acting out is the out force minus the in force?

CurtisB
#4
May11-07, 12:53 AM
P: 16
Pressure Cooker

Well the answer was actually F = Ap_a - (293.15Ap_a)/(393.15), can anyone tell me why the two forces are swapped around, int the first part it was the inside force minus the outside force so why is this the outside force minus the inside force, wouldn't the second part be a negative value because the force is now pushing in from the outside, ie , the direction of the force has changed.
andrevdh
#5
May11-07, 02:19 AM
HW Helper
P: 1,449
Quote Quote by CurtisB View Post
Thanks, that really helped, I got the answer to part one to be 0.34Ap_a ......
How did you get to this answer?
CurtisB
#6
May11-07, 03:03 AM
P: 16
if [tex]pV=Nk_BT[/tex] then [tex]\frac{p}{T} = \frac{Nk_B}{V} = constant[/tex]


so [tex]\frac{p_f}{T_f}=\frac{p_i}{T_i}[/tex], convert [tex]^oC[/tex] to K so [tex]\frac{p_f}{393.15}=\frac{p_i}{293.15}[/tex]


and because [tex]p=\frac{F}{A}[/tex] and [tex]p_i = p_a[/tex],


[tex]\frac{F}{A}=\frac{393.15p_a}{293.15}[/tex] so [tex]F=\frac{393.15AP_a}{293.15}=1.34Ap_a[/tex]


and since the force on the outside acting in = [tex]Ap_a[/tex] then


[tex]1.34Ap_a - Ap_a = 0.34Ap_a[/tex]
andrevdh
#7
May11-07, 03:40 AM
HW Helper
P: 1,449
Ok, now I am on the same page as you.

Quote Quote by CurtisB View Post
Well the answer was actually F = Ap_a - (293.15Ap_a)/(393.15), can anyone tell me why the two forces are swapped around, int the first part it was the inside force minus the outside force so why is this the outside force minus the inside force, wouldn't the second part be a negative value because the force is now pushing in from the outside, ie , the direction of the force has changed.

The problem just wants the resultant force on the top, not taking the direction of the force into consideration. The inner pressure is now lower than the outer pressure. So the resultant force will be inwards and its magnitude is the outer force minus the inner force.
get_physical
#8
Jun15-08, 06:53 PM
P: 85
Quote Quote by CurtisB View Post


and since the force on the outside acting in = [tex]Ap_a[/tex] then
How do we know that the force outside acting in is A*P_a?
Isn't it just A since p(outside)= 1atm ?


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