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stepan [7]
3 years ago
13

Tres litros de oxigeno gaseoso a 15 grados centígrados y a presión atmosférica (1atm), se lleva a una presión de 10mm de Hg. ¿ c

ual será ahora el volumen que ocupe el gas si la temperatura no ha variado?
Physics
1 answer:
lana [24]3 years ago
8 0

Answer:

Three liters of gaseous oxygen at 15 degrees Celsius and at atmospheric pressure (1atm), is brought to a pressure of 10mm Hg. What will be the volume of the gas now if the temperature has not changed?

Explanation:

Given that, the temperature is constant

Then, using gay Lussac law

P1•V1 / T1 = P2•V2 / T2

Since temperature is constant

Then, T1 = T2 = T and they cancels out

So, we are left with

P1•V1 = P2•V2

Given that, .

Initial volume

V1 = 3 litres.

Initial pressure

P1 = 1atm = 101325 Pa

Final pressure

P2 = 10mmHg = 1333.22 Pa

Then, we want to find the final volume V2

Make V2 subject of formula.

V2 = V1•P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litres

So, the final volume is 288 litres.

In Spanish

Dado que la temperatura es constante

Luego, usando la ley gay de Lussac

P1 • V1 / T1 = P2 • V2 / T2

Como la temperatura es constante

Entonces, T1 = T2 = T y se cancelan

Entonces, nos quedamos con

P1 • V1 = P2 • V2

Dado que, .

Volumen inicial

V1 = 3 litros.

Presión inicial

P1 = 1atm = 101325 Pa

Presión final

P2 = 10 mmHg = 1333.22 Pa

Entonces, queremos encontrar el volumen final V2

Hacer V2 sujeto de fórmula.

V2 = V1 • P1 / P2

V2 = 3 × 101325 / 1333.22

V2 = 288 litros

Entonces, el volumen final es de 288 litros

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You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
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Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

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v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

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