Question

(1). The vapor pressure of liquid heptane, C₇H₁₆, is 100. mm Hg at 315 K.

A sample of C₇H₁₆ is placed in a closed, evacuated 522 mL container at a temperature of 315 K. It is found that all of the C₇H₁₆ is in the vapor phase and that the pressure is 73.0 mm Hg. If the volume of the container is reduced to 369 mL at constant temperature, which of the following statements are correct?

Choose all that apply.

Only heptane vapor will be present.

Liquid heptane will be present.

Some of the vapor initially present will condense.

The pressure in the container will be 103 mm Hg.

No condensation will occur.

(2). The vapor pressure of liquid heptane, C₇H₁₆, is 40.0 mm Hg at 295 K.

A sample of C₇H₁₆ is placed in a closed, evacuated container of constant volume at a temperature of 440 K. It is found that all of the C₇H₁₆ is in the vapor phase and that the pressure is 66.0 mm Hg. If the temperature in the container is reduced to 295 K, which of the following statements are correct?

Choose all that apply.

Liquid heptane will be present.

The pressure in the container will be 44.3 mm Hg.

Only heptane vapor will be present.

No condensation will occur.

Some of the vapor initially present will condense.

Answer 1

Correct option:

Liquid heptane will be present.

Some of the vapor initially present will condense.

The pressure in the container will be 103 mm Hg.

Explanation:

Initial pressure, P₁ = 73.0 mmHg

Initial volume, V₁ = 522 mL

Final volume, V₂ = 369 mL

Let us say that the final pressure = P₂

From Boyle's law,

P₁V₁ = P₂V₂

or, 73.0 x 522 = P₂ x 369

or, P₂ = 103 mmHg

The final pressure inside the container is greater than the vapor pressure of of heptane (100 mmHg) at 315 K.

Therefore, some of the vapor of heptane initially present will condense to for liquid heptane.

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Correct option: Some of the vapor initially present will condense.

Liquid heptane will be present.

The pressure in the container will be 44.3 mm Hg.

Explanation:

Initial pressure, P₁ = 66.0 mmHg

Initial temperature, T₁ = 440 K

Final temperature, T₂ = 295 K

Let us say that the final pressure = P₂

Now, from Gay-Lussac's law,

P₁/T₁ = P₂/T₂

or, 66.0/440 = P₂/295

or, P₂ = 44.3 mmHg

The final pressure inside the container is greater than the vapor pressure of of heptane (40.0 mmHg) at 295 K.

Therefore, some of the vapor of heptane initially present will condense to for liquid heptane.