Calculating Gas Pressure in a Piston System with Added Weight

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SUMMARY

The discussion focuses on calculating gas pressure in a piston system when a 0.5 kg weight is added. Initially, the gas pressure is 1 atmosphere at 300 Kelvin. The pressure exerted by the weight is calculated using the formula \( P = \frac{F}{A} \), where \( F \) is the force due to the weight (4.9 N) and \( A \) is the area of the piston. The volume of gas can be determined using the ideal gas law \( PV = nRT \), but the area of the piston must be known to solve for pressure accurately.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Basic physics concepts of force and pressure
  • Knowledge of units of measurement (e.g., Newtons, atmospheres)
  • Familiarity with piston systems and their mechanics
NEXT STEPS
  • Calculate pressure using \( P = \frac{F}{A} \) with known piston area
  • Explore the relationship between pressure, volume, and temperature in gas systems
  • Learn about the effects of varying weights on gas pressure in piston systems
  • Investigate how to manipulate the ideal gas law for different scenarios
USEFUL FOR

Students studying physics, engineers working with pneumatic systems, and anyone interested in thermodynamics and gas laws.

dragon162
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Homework Statement


Suppose the system is at room temperature, 300 kelvin, and that before we put weights on top of the plunger in the cylinder the pressure inside is 1 atmosphere.

1)If i place a .5kg block on top of the piston, what is the pressure of the gas inside the cylinder?
2)At that pressure what is the volume of the gas?
3)If i want to have L=0cm what should the pressure be?
4)how much mass would we have to pile on top of the piston to achieve that pressure inside the system?

Homework Equations


PV=nRT
Volume of piston is V=L*A


The Attempt at a Solution


Ok I am confused as how to find the pressure of the gas when the block is added to the piston. Since the block is on the piston doesn't it apply a pressure of mg, (.5kg)(9.8)=4.9?

The second part is obvious, just use PV=nRT but I have to find the pressure in part 1 first. As for the other two I don't know how to start them but I am focused on the first part right now.
 
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dragon162 said:

The Attempt at a Solution


Ok I am confused as how to find the pressure of the gas when the block is added to the piston. Since the block is on the piston doesn't it apply a pressure of mg, (.5kg)(9.8)=4.9?
mg would be a force, not a pressure. The block applies a force of mg. To get the pressure, you would need to know the area of the piston's face. Without that information, the problem cannot be solved.
 

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