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ZanzabumX [31]
3 years ago
13

Which option best describes the increase in pressure when the volume of a plastic bottle decreases?

Physics
1 answer:
Allisa [31]3 years ago
6 0

Answer:

B) uniform

Explanation:

We can solve this problem by using Boyle's law, which states that:

<em>For a constant mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume.</em>

Mathematically:

where

p is the pressure of the gas

V is its volume

If we apply the equation to the bottle in the problem, we see that:

- when the volume of the bottle (and therefore, of the gas inside) decreases, than the pressure will increase

- viceversa, when the volume of the bottle increases, the pressure will decrease

The amount by which the pressure increases is inversely proportional to the decrease in volume, so the answer depends on how the volume of the bottle decreases: however, if the volume of the bottle decreases uniformly, then we can say that the pressure inside the bottle will also increases uniformly.

MARK THIS ANSWER AS THE BRAINLIEST PLEASE ❤️

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A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

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v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

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mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

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using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

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