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Anon25 [30]
2 years ago
7

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?
Physics
1 answer:
ValentinkaMS [17]2 years ago
7 0

Answer:

the value of the final pressure is 0.168 atm

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

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Answer:

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b) Check explanatios for b

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A linear equation is of the form y = Gx

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x = \left[\begin{array}{ccc}100\end{array}\right]

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