134k views
0 votes
A cylindrical tank has a tight-fitting piston that allows the volume of the tank to be changed. The tank originally contains air with a volume of 0.175 m^3 at a pressure of 0.350 atm. The piston is slowly pulled out until the volume of the gas is increased to 0.365 m^3.

Required:
If the temperature remains constant, what is the final value of the pressure?

User Ezitoc
by
8.4k points

1 Answer

0 votes

Answer:

the value of the final pressure is 0.168 atm

Step-by-step explanation:

Given the data in the question;

Let p₁ be initial pressure, v₁ be initial volume.

After expansion, p₂ is final pressure and v₂ is final volume.

So using the following equations;

p₁v₁ = nRT

p₂v₂ = nRT

hence, p₁v₁ = p₂v₂

we find p₂

p₂ = p₁v₁ / v₂

given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,

final volume v₂ = 0.365 m³

we substitute

p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³

p₂ = 0.06125 atm-m³ / 0.365 m³

p₂ = 0.168 atm

Therefore, the value of the final pressure is 0.168 atm

User Shreyansp
by
8.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.