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satela [25.4K]
2 years ago
15

Explain the effect of strong intermolecular forces on each of these parameters: (a) critical temperature;

Chemistry
1 answer:
Delicious77 [7]2 years ago
7 0
  • (A) The molecules can no longer be condensed because they are moving too quickly at the critical temperature.
  • Because the forces are insufficient to stop molecular motion, this temperature drops with weaker intermolecular forces.
  • Alternately, the critical temperature rises as intermolecular forces increase.

b) The boiling point rises as intermolecular forces do because it becomes harder and more energy-intensive to separate molecules from the liquid phase.

(C) For the same reason as in b), the vapor pressure falls as intermolecular forces rise.

  • Strong intermolecular forces prevent molecules from entering the vapor phase easily at any given temperature, resulting in a decrease in vapor pressure.

d) The heat of vaporization rises as intermolecular forces increase because more energy is required to separate molecules from one another.

Learn more about critical temperature

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<u>The complete question is -</u>

Explain the effect of strong intermolecular forces on each of these parameters: (a) critical temperature, (b) boiling point, (C ) vapor pressure, (d) heat of vaporization

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Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

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