3 days ago

The gauge pressure at the bottom of a cylinder of liquid is pg = 0.40 atm. The liquid is poured into another cylinder with twice

the radius of the first cylinder. What is the gauge pressure at the bottom of the second cylinder?

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serg [3582]

Answer:

I apologize

Explanation:

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4 0

2 months ago

Yuliya22 [3333]

Answer: $0.10233nm$

Explanation:

The mean free path of an atom can be calculated using the following equation:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is referred to as the Universal gas constant

$T=0\°C=273.115K$ represents the absolute standard temperature

$d$ denotes the diameter of helium atoms

$N_{A}=6.0221(10)^{23}/mol$ symbolizes Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ indicates absolute standard pressure

<pFrom this, we can solve for $d$ using (1), aiming to determine the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius equals half of that diameter:

$r=\frac{d}{2}$ (5)

Eventually:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

Nonetheless, we were tasked with finding this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Ultimately:

$r=0.10233nm$ Represents the radius of the helium atom in nanometers.

5 0

3 months ago