MCAT Physical : Physical Chemistry

Study concepts, example questions & explanations for MCAT Physical

varsity tutors app store varsity tutors android store

Example Questions

Example Question #1 : Thermodynamic Systems And Calorimetry

Which term correctly identifies a system in which there is an exchange of heat with the surrounding environment, but no exchange of matter?

Possible Answers:

Conserved system

Isolated system

Closed system

Open system

Thermodynamic system

Correct answer:

Closed system

Explanation:

In a closed system, energy and work can be exchanged with the surrounding area but matter is contained. Consider, for example, a beaker of ice. Heat from the surroundings it transferred to the ice, causing it to melt, but there is no exchange of matter. No external compounds react with the ice and none of the ice leaves the beaker.

In an open system, both heat and matter can be exchanged. In an isolated system neither heat, nor matter is exchanged.

Example Question #1 : Thermodynamic Systems And Calorimetry

You begin an experiment with  of ice at . How much heat is required to melt the ice and raise its temperature to ?

Possible Answers:

Correct answer:

Explanation:

This process can be divided into three stages: the raising of the temperature of the ice, the melting of the ice into water, and the raising of the temperature of the water. The first and third step use the specific heat of the substance to calculate the heat required, while the second step uses the heat of fusion.

Step 1 and 3:

Step 2:

We know the mass of the water, the change in temperature, and the constant values. Using these values, we can sum the heat required for each step to get our final answer.

Example Question #8 : General Thermodynamics

Use the following values for water as needed.

If burning wood releases  of heat energy per gram of wood consumed, what mass of wood must be consumed to heat  of water from  to , and then to convert it to water vapor?

Possible Answers:

Correct answer:

Explanation:

There are two processes requiring added heat in this problem:

1. Raising the temperature of the liquid water from  to  (use )

2. Boiling the water at a constant temperature of  (use )

To use either of these equation, we need to find the mass of the water using the relation between mass, density, and volume.

Use this mass with the given specific heat and temperatures to find the heat for part 1 of the process.

Then, use the mass with the given heat of vaporization to find the energy needed to convert the water to water vapor.

Sum the energies for step 1 and step 2.

This is the total amount of energy needed from the burning wood. Use stoichiometry to find the grams of wood needed to produce this amount of energy.

Example Question #11 : Thermodynamic Systems And Calorimetry

A metal object that is dropped into water. In which scenario would heat be transferred from the object to the water?

Possible Answers:

Correct answer:

Explanation:

If heat is transferred from the metal object to the water, that means that the temperature of the metal must be greater than that of the water. We must find the answer choice where this is the case. Since the temperatures are given in different units, they must be converted to the same unit for comparison purposes.

Only one answer option results in the metal having a greater temperature than the water:

Example Question #41 : Thermochemistry And Energetics

A 35g piece of aluminum at a temperature of 373K is placed into 150g of water at a temperature of 298K. The aluminum and water eventually become the same temperature. No heat is released to the surroundings.

Water has a specific heat capacity of  and aluminum has a specific heat capacity of .

What is the final temperature of the water and aluminum in the container?

Possible Answers:

301.6K

324.5K

315.5K

308.9K

Correct answer:

301.6K

Explanation:

In this problem, we need to track the transfer of heat from the aluminum to the water. Since the heat acquired by the water is equal to the heat given off by the aluminum, we can set their equations equal to each other; however, in order to avoid a negative number, aluminum's change in temperature will be set as the initial temperature minus the final temperature. This results in the equation below.

 

Example Question #41 : Thermochemistry And Energetics

A 200g sample of gold is subjected to 1.2kJ of heat. 

The specific heat capacity for gold is .

What is the change in temperature as a result of the heating?

Possible Answers:

Correct answer:

Explanation:

When a specific heat capacity is given, we typically use the equation . Since we know all of the factors except for the change in temperature, we can simply solve for .

 

Example Question #4 : General Thermodynamics

Given the enthalpies of formation, what is the enthalpy of combustion of octane in the reaction:

Possible Answers:

Correct answer:

Explanation:

The equation for enthalpy of reaction is:

Given our chemical reaction and the enthalpies of formation, we can find the enthalpy of reaction.

First, find the total enthalpy for the products.

Then, find the total enthalpy for the reactants.

Since the oxygen is elemental, its heat of formation is zero.

Return to the original equation to calculate the final enthalpy of reaction.

Example Question #1 : Enthalpy

The formation of nitrous oxide is a 2-step process.

The overall enthalpy of the reaction is +68kJ.

If the enthalpy change of step 1 is 180kJ, what is the enthalpy change of step 2?

Possible Answers:

More information is needed to answer the question.

–112kJ

248kJ

112kJ

Correct answer:

–112kJ

Explanation:

Hess's law states that the enthalpies of the steps in an overall reaction can be added in order to find the overall enthalpy. Since the overall enthalpy of the reaction is +68kJ, we can use the following equation to find step 2's enthalpy.

Example Question #2 : Enthalpy

Consider the following processes:

1) 

2) 

3) 

Which for the following expresses  for the process  ?

Possible Answers:

Correct answer:

Explanation:

1) 

2) 

3) 

We need to add the given processes in such a way that leaves  on the left side, and  on the right. 

Since there are no  in the final equation, begin by combining processes 1 and 2 to eliminate.

First, reverse process 2 and multiply by two:

2a)

2a)

2a)

Add this result to process 1.

1)

2a)

1+2a) 

Now we need to combine with process 3 to eliminate the . Start by multiplying process 3 by two.

3a)

3a) 

Add this to the combined result from processes 1 and 2.

1+2a)

3a)

1+2a+3a) 

This gives our final combination. The elemental reaction is , and the enthalpy of the process is .

Example Question #1 : Entropy

A scientist prepares an experiment to demonstrate the second law of thermodynamics for a chemistry class. In order to conduct the experiment, the scientist brings the class outside in January and gathers a cup of water and a portable stove.

The temperature outside is –10 degrees Celsius. The scientist asks the students to consider the following when answering his questions:

Gibbs Free Energy Formula:

ΔG = ΔH – TΔS 

 

Liquid-Solid Water Phase Change Reaction:

H2O(l) ⇌ H2O(s) + X

The scientist prepares two scenarios. 

Scenario 1:

The scientist buries the cup of water outside in the snow, returns to the classroom with his class for one hour, and the class then checks on the cup. They find that the water has frozen in the cup.

Scenario 2: 

The scientist then places the frozen cup of water on the stove and starts the gas. The class finds that the water melts quickly.

After the water melts, the scientist asks the students to consider two hypothetical scenarios as a thought experiment. 

Scenario 3:

Once the liquid water at the end of scenario 2 melts completely, the scientist turns off the gas and monitors what happens to the water. Despite being in the cold air, the water never freezes.

Scenario 4:

The scientist takes the frozen water from the end of scenario 1, puts it on the active stove, and the water remains frozen.

After the processes in each of the four scenarios above, what can be said about the entropy of the universe?

Possible Answers:

It increases in scenarios 1 and 3, only.

It increases in scenarios 2 and 4, only.

It increases in scenarios 3 and 4, only.

It increases in all scenarios.

It increases in scenarios 1 and 2, only.

Correct answer:

It increases in scenarios 1 and 2, only.

Explanation:

Scenarios 3 and 4 are presented as thought experiments because, common sense tells us, they will never happen. The fundamental reason they will never happen is because they do not lead to a total increase in entropy of the universe. In order for an event to proceed, this requirement must be satisfied.

Learning Tools by Varsity Tutors