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Andre45 [30]
3 years ago
14

All three phases can exist in an equilibrium with one another at the __ of a substance

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

Triple point

Explanation:

Yes, all three phases can exist in an equilibrium with one another at the triple point of a substance.

In thermodynamics, the triple point of a substance is defined as the point at which all three phases of substance co-exist together at a particular pressure and temperature.

for example, the triple point of mercury occurs at a temperature of −38.83440 °C and a pressure of 0.2 mPa.

And for water it is 0.01 °C and 611.73 pascals.

you will get some clear idea after seeing the graph.

And critical point as you see in the graph is end point of phase equilibrium curve, mainly liquid-vapor equilibrium

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a= 0.2m/s²

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20 = 100a (divide both sides by 100)

a = 0.2m/s²

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When a calcium atom forms an ion, it loses two electrons. what is the eletrical charge ofbthe calcium ion?
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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at
Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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