Thermodynamics - intial pressure

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Homework Help Overview

The discussion revolves around determining the initial pressures in two containers holding a monatomic gas at a constant temperature. Container A is rigid, while Container B has a movable piston. The problem involves understanding the effects of the piston weight and atmospheric pressure on the pressure calculations for both containers.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between pressure, volume, and temperature in both containers, questioning how the weight of the piston influences the pressure in Container B compared to Container A. Some participants suggest using the ideal gas law to relate the pressures, while others focus on the mechanics of the piston system.

Discussion Status

The discussion is active, with participants offering insights into the mechanics of pressure in the context of the piston and atmospheric effects. There is a recognition that the pressures in both containers may not be directly comparable without considering the additional forces acting on the piston.

Contextual Notes

Participants note that the volume of gas in Container B is influenced by the position of the piston, which introduces complexity to the pressure calculations. There is also an acknowledgment of the need for atmospheric pressure to be included in the calculations for Container B.

amdma2003
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Homework Statement



Two 800 [tex]cm^{3}[/tex] containers hold identical amounts of a monatomic gas at 20 deg C. Container A is rigid. Container B has a 100 [tex]cm^{2}[/tex] piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.

What are the initial pressures in containers A and B?
The containers are not heated initially.

Homework Equations



To find the pressure of Container B we can use the equation [tex]P = \frac{F}{A}[/tex] where [tex]F = mg = 10 kg \times 9.8 \frac{m}{s^{2}}[/tex] and [tex]A = 100 cm^{2}[/tex]


The Attempt at a Solution



As I have found the Container B to be at 9800 Pascals (by converting cm to meters) I do not know how to approach the problem for finding the pressure for Container A.
 
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The pressure on a surface is caused by the atoms colliding with it, like a ball rebounding from a wall. This imparts an impulse to the surface - it experiences a force due to the collisions. The strength of these collisions are determined by two factors - the mass and (average) speed of the atoms. The speed of the atoms in both containers are the same since they are at the same temperature and both have atoms with similar mass (the speed would have been different if the mass were not the same, even if the temperature were still the same. That is the temperature is a measure of the [average] kinetic energy of the gas, which depends on both the speed and the mass of the atoms). What can one then conclude about the initial pressure of the gas in container A?
 
Last edited:
amdma2003 said:

Homework Statement



Two 800 [tex]cm^{3}[/tex] containers hold identical amounts of a monatomic gas at 20 deg C. Container A is rigid. Container B has a 100 [tex]cm^{2}[/tex] piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.

What are the initial pressures in containers A and B?
The containers are not heated initially.

Homework Equations



To find the pressure of Container B we can use the equation [tex]P = \frac{F}{A}[/tex] where [tex]F = mg = 10 kg \times 9.8 \frac{m}{s^{2}}[/tex] and [tex]A = 100 cm^{2}[/tex]


The Attempt at a Solution



As I have found the Container B to be at 9800 Pascals (by converting cm to meters) I do not know how to approach the problem for finding the pressure for Container A.

You can determine P in Container B as you have done except that you also have to add in the atmospheric pressure.

Find PV/n = RT for Container B. You know that PV has to be equal for both since n is the same. The difference in pressure between the two is due to the weight of the piston

AM
 
Last edited:
Since V and n is the same for both can't one take your reasoning further and say that

P = nRT/V

should be the same for both?
 
andrevdh said:
Since V and n is the same for both can't one take your reasoning further and say that

P = nRT/V

should be the same for both?
V is not necessarily the same for both. V is the volume of the gas, not the container. The position of the piston determines the volume occupied by the gas in B.
-------------

Upon further consideration, I think you must be right. The volume of the "container" in B must be the 800 cm^3. Otherwise, there is not enough information. P_B = P_A.

AM
 
Last edited:
I don't understand how the weight of the piston affects the pressure. I thought since the piston is free to move up and down, the pressure is just mg/A.
 
amdma2003 said:
I don't understand how the weight of the piston affects the pressure. I thought since the piston is free to move up and down, the pressure is just mg/A.
Do a freebody diagram of the forces on the piston.

There are two forces downward: the piston weight (= mg) + the atmosphere (= [itex]P_{atm}A_{piston}[/itex])

There is one upward force: [itex]P_{B}A_{piston}[/itex]

Since there is no acceleration, they are equal and opposite forces:[tex]P_{B}A_{piston} = mg + P_{atm}A_{piston}[/tex]

[tex]P_{B} = \frac{mg + P_{atm}A_{piston}}{A_{piston}} = \frac{mg}{A_{piston}} + P_{atm}[/tex]

AM
 

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