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Lesechka [4]
3 years ago
11

A sample of oxygen gas has a volume of 165 mL at a pressure of 0.854 atm. What will the volume of the gas be at a pressure of 2.

3 atm if the temperature remains constant?
Physics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer: 24.1 mL

Explanation:

Initial volume V1 = 65 mL

Initial pressure P1 = 0.854 atm

Final volume of the gas V2 = ?

Final pressure of the gas P2 = 2.3 atm

Since, pressure and volume are involved while temperature is constant, apply the formula for Boyle's law

P1V1 = P2V2

0.854 atm x 65 mL = 2.3 atm x V2

55.51 atm mL = 2.3 atm x V2

V2 = (55.51 atm mL / 2.3 atm)

V2 = 24.1 mL

Thus, the final volume of the gas will be

24.1 mL

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Solution :

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$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$

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r - 1 = 7130

r = 7130 + 1

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$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$

$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$

$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$

$v=0.99999999017C$

Δ = 1 - 0.99999999017

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Relative mass, $m_{rel}=r.m$

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As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
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Answer:

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F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N

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