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3 years ago

HURRY PLEASE

Chemistry

2 answers:

katrin2010 [14]3 years ago

4 0

Answers:

Question 1:

The diagram for gallium will have flat, horizontal lines at 30 and 2204 °C.

Questoin 2:

The diagram for methane will have a diagonal line representing the liquid phase between -183°C and -162°C.

Question 3:

For gold, the boiling point corresponds to the y-value at 2856 °C of the top horizontal line

Question 4.

For nitrogen, the line at -210°C will be flat, which represents the change from a solid to a liquid

Explanations:

Question 1:

The diagram for gallium will have flat, horizontal lines at 30 and 2204 °C.

The table shows that the melting point of gallium is 30°C and its boiling point is 2204°C.

Melting point is the temperature at which the substace changes its state from solid to liquid. During that change, the temperature of the substance does not change, because the heat supplied is used to accomplish the phase change. So, the temperature is constant and that means that portion of the diagram is flat.

The same is valid during boiling: the temperature remains constant while the substance is passing from liquid to gas at the boiling point.

Questoin 2:

The diagram for methane will have a diagonal line representing the liquid phase between -183°C and -162°C.

Between the melting (-183°C) and boiling (-162°C) points of methane, its temperature will increase more or less linearly, which is represented with a diagonal (slant) line between those points. During this interval the heat is used to increase the temperature and no phase of change happens.

Question 3:

For gold, the boiling point corresponds to the y-value at 2856 °C of the top horizontal line

The table shows that the boiling point of gold is 2,856°C.

In a temperature-vs.-time diagram the temperature is represented on the vertical axis (y-value) and the time is represented on the horizontal axis.

Since, the temperature of the substance does not change during boiling, the line during the time that this change of phase is happening is flat. And since this temperatue is higher than the melting temperature, this is the top horizontal line in the diagram.

Question 4.

For nitrogen, the line at -210°C will be flat, which represents the change from a solid to a liquid

The table shows that the melting point of nitrogen is -210°C, that means that the temperature will remain constant at -210°C while the substance is absorbing heat to pass from solid to liquid.

In conclusion, you must remember that all the phase changes, melting (from solid to liquid), freezing (from liquid to solid), boilng (from liquid to gas), and condensing (from gas to liquid) happens at constant temperature, and so the temperature - vs. - time diagrams show flat lines (constant y-values) during those intervals of time.

LenaWriter [7]3 years ago

4 0

Answer:

At 30 and 2,204.

Diagonal and liquid phase.

2,856 and top horizontal line.

Flat and the change from a solid to a liquid.

Explanation:

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Explanation:

Assuming ideal gas behaviour of air , then

P*V= n*R*T = m / M * R *T

since

ρ= m/V = P*M /( R *T)

where

n= moles , V= volume , m= mass

ρ= density

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M= molecular weight of air = 0.21*32+0.79*28= 28.24 gr/mol = 0.02824 kg/mol

T= absolute temperature = 10°C + 273 = 283 K

R= ideal gas constant = 8.314 J/mol K

solving for ρ

ρ= P*M /( R *T) = 120000 Pa*0.02824 kg/mol/(8.314 J/mol K*283 K) = 1.47 kg/m³

then from Bernoulli's equation

P₁ + ρ*v₁²/2 = P₂ + ρ*v₂²/2

where 1 denotes inlet and 2 denotes other point , p = pressure and v= velocity . Then solving for p₂

P₁ + ρ*v₁²/2 = P₂ +ρ*P₂²/2

P₂= P₁ +ρ*v₁²/2 - ρ*v₂²/2 = P₁ +ρ/2*(v₁² - v₂²)

replacing values

P₂= P₁ +ρ/2*(v₁² - v₂²) = 120000 Pa + 1.47 kg/m³/2*[(30 m/s)²-(250 m/s)²] = 74724 Pa = 74 kPa

P₂= 74 kPa

then if the temperature remains constant

ρ₁= P₁*M /( R *T) and ρ₂= P₂*M /( R *T)

dividing both equations

ρ₂/ρ₁ = P₂/ P₁

ρ₂ = (P₂/ P₁)*ρ₁

then from Bernoulli's equation

P₁ + ρ₁*v₁²/2 = P₂ + ρ₂*v₂²/2

P₂ = P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2

therefore

ρ₂ = (P₂/ P₁)*ρ₁ = (P₁ + ρ₁*v₁²/2 - ρ₂*v₂²/2 ) /P₁ *ρ₁

P₁ * ρ₂ = P₁ *ρ₁ + ρ₁²*v₁²/2 - ρ₂*ρ₁ * v₂²/2

P₁ * ρ₂ + ρ₂*ρ₁ * v₂²/2 = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂* (P₁ + ρ₁ * v₂²/2) = P₁ *ρ₁ + ρ₁²*v₁²/2

ρ₂ = (P₁ *ρ₁ + ρ₁²*v₁²/2)/(P₁ + ρ₁ * v₂²/2) = (P₁ + ρ₁*v₁²/2)/(P₁/ρ₁ + v₂²/2)

replacing values

ρ₂ = ( 120000 Pa + 1.47 kg/m³/2*(30 m/s)²)/(120000 Pa/1.47 kg/m³+1/2*(250 m/s)²)

ρ₂ = 1.06 kg/m³

the error of assuming constant ρ would be

e = (ρ₂ - ρ)/ρ₂= 1- ρ/ρ₂= 1- 1.47 kg/m³/1.06 kg/m³ = -0.386 (-38.6%)

thus Bernoulli’s equation should not be applied

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