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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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The right choice is:
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3 years ago
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A person pushes a 15.7-kg shopping cart at a constant velocity for a distance of 25.9 m on a flat horizontal surface. She pushes
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Answer:

a)    F = 35.7 N, b)   W = 846.7 J, c)   W = - 846.9 J, d) W=0

Explanation:

a) For this exercise let's use Newton's second law, let's set a reference frame with the x-axis horizontally

let's break down the pushing force.

        cos (-23,7) = Fₓ / F

        sin (-237) = F_y / F

        Fₓ = F cos 23.7 = F 0.916

        F_y = F sin (-23.7) = - F 0.402

         

Y axis  

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       N = W + F 0.402

X axis

       Fₓ - fr = 0

       F 0.916 = fr

       F = fr / 0.916

       F = 32.7 / 0.916

       F = 35.7 N

It is asked to calculate several jobs

b) the work of the pushing force

       W = fx x

       W = 35.7 cos 23.7 25.9

       W = 846.7 J

c) friction force work

        W = F x cos tea

friction force opposes movement

        W = - fr x

         W = - 32.7 25.9

         W = - 846.9 J

d) The work of the force would gravitate, as the displacement and the force of gravity are at 90º, the work is zero

          W = 0

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2 years ago
A 0.2 kg plastic cart and a 20 kg lead cart both roll without friction on a horizontal surface. Equal forces are used to push bo
Fynjy0 [20]

Answer:

Equal

Explanation:

The impulse theorem states that the impulse exerted on each cart is equal to the change in momentum of the cart:

I=\Delta p=p_f -p_i

where

I is the impulse

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pi is the initial momentum

The impulse is equal to the product between the force applied and the contact time:

I=F\Delta t

In this case, the force applied to the two carts (F) is the same, and the contact time (\Delta t) is the same as well. Therefore, the impulse exerted on the two carts is the same.

Moreover, the initial momentum of the two carts is also the same (zero, because they start from rest: p_i=0). So the formula becomes

p_f = I

And since I is the same for the two carts, the final momentum (p_f) will also be equal.

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Answer:

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