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BaLLatris [955]
4 years ago
5

For each of the following pairs of gas properties, describe the relationship between the properties, describe a simple system th

at could be used to demonstrate the relationship, and explain the reason for the relationship: (a) volume and pressure when number of gas particles and temperature are constant, (b) pressure and temperature when volume and the number of gas particles are constant, (c) volume and temperature when pressure and the number of gas particles are constant, (d) the number of gas particles and pressure when volume and temperature are constant, and (e) the number of gas particles and volume when pressure and temperature are constant.
Physics
1 answer:
asambeis [7]4 years ago
7 0

Answer:

For  each pair of properties of a gas, the relationships are (see the explanation for the description of the systems):

  1. (a) Volume and pressure: The relationship between them is inversely proportional.
  2. (b) Pressure and temperature: They have a directly proportional relationship.
  3. (c) Volume and temperature: They relationship is directly proportional.
  4. (d) Number of gas particles and pressure: The relationship is directly proportional between them.

Explanation:

1. Volume and pressure (temperature and amount of particles constant):

They have an inversely proportional relationship, because <em>if volume is reduced, the pressure increases, or if the volume increases, the pressure decreases</em>.

A simple system could be one similar to the one used by Boyle to test this relationship:

  • Seal the short extreme of a translucent J tube. It could be glass or plastic.
  • Put some water on it. As much as needed to have both sides of the tube filled.
  • Using a syringe, and a flexible small tube,inject a determined volume of air in the bottom in a way that the bubble is trapped in the seal side of the J tube.
  • Then if more water is added to the tube, it will increase the pressure (from the pressure definition is possible to in the trapped air, and is possible to measure the compression of the air bubble. The same is possible if using the syringe, and the flexible tube, some water is removed, and the increasing of volume could be observed.

2. Pressure and temperature (volume and amount of particles of the gas remains constant)

They have a directly proportional relationship, because <em>if temperature is reduced, the pressure decreases, or if the temperature increases, the pressure would increase, also</em>.

A simple system to show this is two cans of soda.

  • The can is rigid, so the volume is always constant, and the amount of gas inside the soda is the same.
  • Put one can under the sun, and the other in the cooler.
  • After a while, take it out the can in the cooler, and open both cans.
  • The one that was under the sun will "explode", in other words, it will liberate a lot of foam of gas and soda, meaning that the pressure inside the can was high.
  • The one that was in te cooler, won't liberate any foam, meaning that the pressure was low.

3. Volume and temperature (pressure and amount of particles of the gas remains constant)

They have a directly proportional relationship, because <em>if temperature is reduced, the pressure decreases, or if the temperature increases, the pressure will increase, also</em>.

A simple system to show this is a party balloon.

  • Fill the party balloon with some air, not enough to be close to explode, but enough to have it of a medium size. Tie the filling hole of the balloon.
  • The air inside the balloon would be at the same pressure than the atmosphere around it, so always will be at this pressure, and the close hole ensure that it has always the same amount of air inside.
  • Now is possible to use some heat source, for example as a hair dryer to increase the temperature of the balloon and its contents. The size of the balloon will increase. Then using water is possible to cool it down and watch how its size decreases.

4. Number of gas particles and pressure (volume and temperature of gas remains constant)

They have a directly proportional relationship, because <em>if the amount of gas particles is reduced, the pressure decreases, or if quantity of gas particles increases, the pressure will increase, also</em>.

A simple system to show this would be a bicycle tire:

  • The tire is rigid, so its volume is essentially constant, and the temperature would remains the same if not moving or driving it.
  • Using a tire gauge, it is possible to know the manometric pressure inside the tire, that is the difference between the actual pressure inside the tire and the atmospheric pressure.
  • Then each time that using an air pump some air is injected in the tire, it si possible to check the pressure inside it using the gauge, and observe how is increasing.
  • Also, is possible to open the valve, to allow some air to escape, then use the gauge to observe how the pressure decreases.

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Answer:

741 J/kg°C

Explanation:

Given that

Initial temperature of glass, T(g) = 72° C

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Temperature of liquid, T(l)= 40° C

Final temperature, T(2) = 57° C

Specific heat capacity of the liquid, c(l) = ?

Using the relation

Heat gained by the liquid = Heat lost by the glass

m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)

Since their mass are the same, then

C(l)ΔT(l) = C(g)ΔT(g)

C(l) = C(g)ΔT(g) / ΔT(l)

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Calculate the moment of inertia for each scenario: (a) An 80.0 kg skater is approximated as a cylinder with a 0.140 m radius. (b
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Answer:

a) the moment of inertia is 0.784 Kg*m²

b) the moment of inertia is with arms extended is 1.187 Kg*m²

c) the angular velocity in scenario (b) is 4.45 rad/s

Explanation:

The moment of inertia is calculated as

I = ∫ r² dm

since

I = Ix + Iy

and since the cylinder rotates around the y-axis then Iy=0 and

I = Ix = ∫ x² dm

if we assume the cylinder has constant density then

m = ρ * V = ρ * π R²*L = ρ * π x²*L

therefore

dm = 2ρπL* x dx

and

I = ∫ x² dm = ∫ x² 2ρπL* x dx = 2ρπL∫ x³ dx = 2ρπL (R⁴/4 - 0⁴/4) = ρπL R⁴ /2 =  mR² /2

therefore

I skater = mR² /2 = 80 Kg * (0.140m)²/2 = 0.784 Kg*m²

b) since the arms can be seen as a thin rod

m = ρ * V = ρ * π R²*L = ρ * π R²*x

dm =ρ * π R² dx

I1 = ∫ x² dm = ∫ x² * ρ * π R² dx = ρ * π R²*∫ x² dx = ρ * π R²* ((L/2)³/3 - (-L/2)³/3)

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therefore

I skater 2 = I1 + I skater =  mL²/12 + mR² /2= 8 Kg* (0.85m)²/12 +(80-8) Kg * (0.140m)²/2 = 1.187 Kg*m²

c)  from angular momentum conservation

I s2 * ω s2 = I s1 * ω s1

thus

ω s2 = (I s1 / I s2 )* ω s1 /= (0.784 Kg*m²/1.187 Kg*m²) * 6.75 rad/s = 4.45 rad/s

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Answer:

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