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atroni [7]
3 years ago
13

Which of the following statements are evidence that gases do not always behave ideally? Check all that apply.?

Physics
2 answers:
yaroslaw [1]3 years ago
8 0

Your answer will be A

Hope this helps you out a little


Lelechka [254]3 years ago
8 0

You can choose :

A) CO2 gas becomes dry ice (solid CO2) at 1 atm and –78.5 °C.

B) When two gases mix, Dalton's law of partial pressure follows.

C) At 4 K and 1 atm, helium is a liquid.

<em>Because all of them is the right answer.</em>

<h2>Further explanation</h2>

Gas is a substance that is composed of atoms or molecules (commonly called particles) with relatively far distances. Between gas particles, there are weak bonds, so weak are these bonds that the physical properties of substances are almost [1] not affected by the type of substance. For example fluorine and helium. Chemically the two are very different, but their physical behavior is very similar. Various studies conducted in the 1600s found that the physical properties of all gases were determined by:

  • Pressure (P)
  • Volume (V)
  • Number of moles (n)
  • Temperature (T)

The relationship between these variables is called the gas laws, and the gas that follows these laws is called the ideal gas. There are various kinds of gas laws, but for this discussion, we will discuss the ideal gas law and partial pressures and Dalton's law.

The Ideal Gas Law

When Avogadro's law, Charles's law, and Boyle's law combine to produce a statement called the ideal gas, which states that the volume of a gas is influenced by the number of moles, pressure, and temperature. Ideal gas shows that there must be gas-gas that have non-ideal behavior. There is not a single gas that follows the equation perfectly under all conditions. All real gases are non-ideal and deviate from the behavior predicted by equations.

Partial Pressure and Dalton's Law.

The ideal gas law applies not only to pure gas but also to gas mixtures. Gas has a very unconcerned nature to neighboring molecules so that the presence or absence of other gases is unchanged. It can be said, the pressure caused by gas when occupying a container alone is as large as the pressure caused by the gas when occupying the gas together with other gases. The total pressure is the amount of pressure from all gases.

Learn more

Gas laws brainly.com/question/1324262, brainly.com/question/6534032

Details

Class: High School

Subject: Physics

Keyword: Gas laws.

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A horizontal spring with stiffness 0.4 N/m has a relaxed length of 11 cm (0.11 m). A mass of 21 grams (0.021 kg) is attached and
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Answer:

0.6983 m/s

Explanation:

k = spring constant of the spring = 0.4 N/m

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x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m

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Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
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Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

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Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

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La velocidad lineal de la lavadora es:

v = \frac{2\pi\cdot r}{T} (3)

(r = 0.4\,m, T = 0.25\,s)

v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

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\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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