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4 years ago

A gas occupies 2.22 l at 3.67 atm. what is the volume at 1.94 atm?

1 answer:

8 0

According to Boyle's law, volume is inversely proportional to pressure. thus P=k/V
Therefore PV=k
P1V1=P2V2
In the question above,
P1=3.67atm
P2=1.94atm
V1=2.22L
V2=?
Thus substituting for the values in the gas equation;
3.67atm*2.22L=1.94atm*V2
V2=3.67atm*2.22L/1.94atm
=4.21L

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(A) $7.28\cdot 10^{-10} m$

The De Broglie wavelength of an electron is given by

$\lambda=\frac{h}{p}$ (1)

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The electron in this problem has a speed of

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And substituting into (1), we find its De Broglie wavelength

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In this case we have:

m = 0.143 kg is the mass of the ball

v = 40.0 m/s is the speed of the ball

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$p=mv=(0.143 kg)(40.0 m/s)=5.72 kg m/s$

And so, the De Broglie wavelength of the ball is given by

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$y=0.492 cm = 4.92\cdot 10^{-3} m$

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Solving the formula for $\lambda$ , we find the wavelength of the electrons in the beam:

$\lambda=\frac{ya}{L}=\frac{(4.92\cdot 10^{-3}m)(2.00\cdot 10^{-6} m)}{1.091 m}=9.02\cdot 10^{-9}m$

(D) $7.35\cdot 10^{-26}kg m/s$

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$p=\frac{h}{\lambda}$

where here we have

$\lambda=9.02\cdot 10^{-9}m$ is the wavelength

Substituting into the formula, we find

$p=\frac{6.63\cdot 10^{-34}Js}{9.02\cdot 10^{-9}m}=7.35\cdot 10^{-26}kg m/s$

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