Boyle's Law and Applied pressure

AI Thread Summary
In a Boyle's Law Apparatus, the pressure on the plunger can be determined using the formula P = F/A, where F is the force from the mass on the plunger and A is the area. The relationship between initial and final pressure and volume is expressed by Boyle's Law as P1V1 = P2V2. There is a discussion on whether the pressure applied to the plunger (P2) should equal the calculated pressure from P = F/A, with findings indicating they are close but not identical. It is noted that atmospheric pressure must be included in calculations, leading to the equation P2 = F/A + atmospheric pressure. The conversation concludes that under constant temperature conditions, these relationships hold true.
nmsurobert
Messages
288
Reaction score
36
When using a Boyle's Law Apparatus, pressure applied to the plunger can be calculated by knowing the mass of the object on the plunger and the area of the plunger. P = F/A.

The change of pressure inside the cylinder can be calculated using Boyles Law, P1V1 = P2V2

Should the value for the pressure applied to the top of the plunger be the same as P2, the pressure of the gas in the cylinder once the plunger has been compressed? should P2 = F/A ?

I did some math and math shows that they aren't equal. they're close, but not the same. however, i don't know if they should be equal under ideal conditions. using P = F/A, that value is slightly larger than using P2 = (P1V1)/V2
 
Physics news on Phys.org
nmsurobert said:
Summary: P = F/A and Boyles Law relationship

When using a Boyle's Law Apparatus, pressure applied to the plunger can be calculated by knowing the mass of the object on the plunger and the area of the plunger. P = F/A.

The change of pressure inside the cylinder can be calculated using Boyles Law, P1V1 = P2V2

Should the value for the pressure applied to the top of the plunger be the same as P2, the pressure of the gas in the cylinder once the plunger has been compressed? should P2 = F/A ?

I did some math and math shows that they aren't equal. they're close, but not the same. however, i don't know if they should be equal under ideal conditions. using P = F/A, that value is slightly larger than using P2 = (P1V1)/V2
Neglecting the weight of the plunger, if the temperature is constant, the pressure P2 should equal the weight.
 
Chestermiller said:
Neglecting the weight of the plunger, if the temperature is constant, the pressure P2 should equal the weight.

the weight? numerically, P2 = mg?
 
nmsurobert said:
the weight? numerically, P2 = mg?
Plus the atmospheric pressure. In Boyle's law, one must use the absolute pressure.
 
Chestermiller said:
Plus the atmospheric pressure. In Boyle's law, one must use the absolute pressure.

I'm taking atmospheric pressure into account. once a mass has been added to the top of the plunger,
P2 = F/A + 101325 N/m2 = (P1V1)/V2?
 
nmsurobert said:
I'm taking atmospheric pressure into account. once a mass has been added to the top of the plunger,
P2 = F/A + 101325 N/m2 = (P1V1)/V2?
Yes, assuming the temperature is constant.
 
Chestermiller said:
Yes, assuming the temperature is constant.
awesome. thank you!
 

Similar threads

Pressure vessel exploding within an atmospheric chamber
Replies
3
Views
2K
Confused about Boyle's Law, Charles' Law and the pressure Law?
Replies
5
Views
2K
Suction force and pressure difference
Replies
1
Views
3K
Equation for spring force for a cylinder on compressed air
Replies
1
Views
4K
A Calculate the work done by pressure rupturing a spherical containment
Replies
7
Views
1K
Calculation of pressure & volume both isothermally & adiabatically
Replies
8
Views
1K
Volumetric flow rate of nitrogen through relieving pressure regulator
Replies
3
Views
2K
Can a Self-Gravitating Gas Ball Simulate Stellar Formation Dynamics?
Replies
5
Views
2K
How to calculate the leak rate based on pressure drop
Replies
4
Views
5K
What is pressure when there are no container walls?
2
Replies
67
Views
8K

Hot Threads

Recent Insights

Back
Top