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harkovskaia [24]
3 years ago
12

To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?

Physics
1 answer:
Lena [83]3 years ago
7 0

The volume of the gas at 500^{\circ} \text{C} is \fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}.

Further Explanation:

Consider the pressure of the gas to be constant.  

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

\fbox{\begin\\V\propto T\\\end{minispace}}

The above expression can we written as.

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Convert the temperature of the gas into kelvin.

T=273+T^\circ\text{C}}}

Here, T is the temperature in kelvin and T^\circ\text{C}}} is the temperature in centigrade.  

The initial temperature of the gas is 50^\circ\text{C}. The temperature of the gas in kelvin is.  

\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}

The final temperature of the gas is 500^\circ\text{C} . The temperature in kelvin is.  

\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}

Substitute the values of temperature and volume in the expression of the Charles' Law.  

\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}

Thus, the volume of the gas at 500^\circ\text{C}} will be \fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}.

Learn More:  

1. Examples of wind and solar energy brainly.com/question/1062501

2. Stress developed in a wire brainly.com/question/12985068

Answer Details:  

Grade: High school  

Subject: Physics  

Chapter: Gas law  

Keywords:  

Charles law, temperature, volume, initial, final, kelvin, centigrade, 50 C, 500 degree, 500 C, 50 degree, 2,33 L, gas expand, sample, heated.

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Sophie [7]

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• The net force in the parallel direction is

∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

• The net force in the perpendicular direction is

∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0

Solving the second equation for <em>n</em> gives

<em>n</em> = <em>mg</em> cos(21°)

<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)

<em>n</em> ≈ 1.83 N

Then the magnitude of friction is

<em>f</em> = <em>µn</em>

<em>f</em> = 0.25 (1.83 N)

<em>f</em> ≈ 0.457 N

Solve for the acceleration <em>a</em> :

-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

<em>a</em> ≈ -5.80 m/s²

so the block is decelerating with magnitude

<em>a</em> = 5.80 m/s²

down the ramp.

5 0
3 years ago
Which of the following occurs when the fight-or-flight response is triggered?
kiruha [24]

Answer:

<h2>A  or B</h2>

Explanation:

The autonomic nervous system has two components, the sympathetic nervous system and the parasympathetic nervous system. The sympathetic nervous system functions like a gas pedal in a car. It triggers the fight-or-flight response, providing the body with a burst of energy so that it can respond to perceived dangers.

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3 years ago
An astronaut on an alien planet drops a rock into a crater which is 100 meters deep. The rock hits the bottom of the crater 4 se
Semmy [17]

Answer:

The gravity on this planet is stronger than that of earth.

Explanation:

First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:

h = Vi t + (0.5)gt²

where,

h = height or depth of crater = 100 m

Vi = Initial Velocity of rock = 0 m/s

t = time = 4 s

g = acceleration due to gravity on this planet = ?

Therefore,

100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²

g = (200 m)/(16 s²)

g = 12.5 m/s²

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ge = 9.8 m/s²

Since,

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<u>The gravity on this planet is stronger than that of earth.</u>

6 0
3 years ago
Jane has a mass of 55 kg and his body covers 700 nails with a surface area of 1.00 mm,
Oksana_A [137]

We have,

  • Jane mass is 55 kg
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We know that,

  • Pressure = Force/Area

Let's calculate force as we already have area;

  • F = ma
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  • F = 539 N

Now, if should she would be on 700 nails then pressure will be;

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And if should would be on a 1 nail only,

  • P = F/A
  • P = 539/1 × 1000000
  • P = 539000000 Pascal

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6 0
2 years ago
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adell [148]

Answer:

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Explanation:

For this exercise we use Coulomb's law

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In this case they indicate that the load is of equal magnitude

       q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

       F = k \ \frac{q^2}{r^2}

       q = \sqrt{\frac{F \ r^2}{k} }

we calculate

        q = \sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9}  }

        q = \sqrt{7 \ 10^{-12} }Ra 7 10-12

        q = 2.65 10⁻⁶ C

7 0
3 years ago
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