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djverab [1.8K]
3 years ago
12

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

air. Assume the pressure is 1 atm and the temperature is 20∘C. Only about 21% of air is O2. 1. How many oxygen molecules do you inhale if you are at sea level?
Physics
1 answer:
Pavel [41]3 years ago
7 0

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

PV=nRT

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

V = 21\% * 0.66L

V = 0.1386L

Our values are given as

P = 1atm

T=293K R = 0.083145kJ*mol^{-1}K^{-1}

Using the equation to find n, we have:

PV=nRT

n = \frac{PV}{RT}

n = \frac{(1)(0.1386)}{(0.0821)(293)}

n = 5.761*10^{-3}mol

Number of molecules would be found through Avogadro number, then

\#Molecules = 5.761*10^{-3}*6.022*10^{23}

\#Molecules = 3.469*10^{21} molecules

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Answer:

Explanation:

Given

Displacement is \frac{1}{3} of Amplitude

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U=\frac{1}{2}kx^2

U=\frac{1}{2}k(\frac{A}{3})^2

U=\frac{1}{18}kA^2

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Total Energy=kinetic Energy+Potential Energy

K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2

K.E.=\frac{8}{18}kA^2

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P.E.=\frac{1}{4}kA^2

\frac{1}{2}kx^2=\frac{1}{4}kA^2

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Answer:

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n = c/v.

v is the velocity in the medium  (2.3 × 10⁸ m/s)

c is the speed of light in air = 3.0 × 10⁸ m/s

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n = 3.0 × 10⁸ /  2.3 × 10⁸

n = 1.31

Using Snell's law as:

n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}

Where,  

{\theta_i}  is the angle of incidence  ( 25.0° )

{\theta_r} is the angle of refraction  ( ? )

{n_r} is the refractive index of the refraction medium  (air, n=1)

{n_i} is the refractive index of the incidence medium (glass, n=1.31)

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