GRE Subject Test: Chemistry : Physical Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

varsity tutors app store varsity tutors android store

Example Questions

Example Question #1 : Phases Of Matter

Oil and vinegar is a very popular salad dressing. It also is commonly used for dipping bread. The main component of the oil phase is olive oil, while the main part of the vinegar portion is aqueous acetic acid.

When oil and vinegar salad dressing is allowed to stand at room temperature, two distinct phases are observed. The main explanation for this phenomenon is the difference in _________ of the two phases.

Possible Answers:

acidity

polarity

solubility

density

buoyancy

Correct answer:

polarity

Explanation:

Olive oil, like most oils, is non-polar, while aqueous acetic acid is very polar. These two phases do not mix because of their different solvent polarities.

Example Question #22 : Physical Chemistry

A gas sample is contained in a 4L vessel at a pressure of 3atm. Assuming all other conditions are kept constant, what is the new pressure in the vessel if the volume is reduced to 1.5L?

Possible Answers:

Correct answer:

Explanation:

According to Boyle's law, pressure and volume are inversely proprotional to each other. This is represented by the equation:

In other words, as volume decreases in a vessel, the pressure will increase, and vice versa. Using the given conditions, we can solve for the final pressure in the vessel:

Example Question #1 : Gas Laws

An unknown amount of neon gas is contained in a 3.00L vessel. At a temperature of , the gas exerts a pressure of 4.00atm.

Neon gas has a molar mass of .

Based on these conditions, what is the mass of neon gas in the vessel?

Possible Answers:

Correct answer:

Explanation:

This question deals with the amount of gas present in a vessel for only one set of conditions. This makes the ideal gas law a suitable equation to use in order to determine the amount of gas in the vessel. The ideal gas law is written as:

Using this equation, we can solve for the molar quantity of gas in the vessel:

Knowing this, we can now solve for the mass of the gas in the vessel by multiplying this molar amount by the molar mass:

 

Example Question #1 : Phases Of Matter

Which gas follows the exact definition of the ideal gas law?

Possible Answers:

None of these

Correct answer:

None of these

Explanation:

Though the ideal gas law gives a nearly close to real approximation of numbers, it oversimplifies its description of gases. No real gas follows the exact definition of the ideal gas law and is very complex because there are intermolecular forces that must be considered. An ideal gas described as a point mass in which the particles are so small that its volume is negligible. However, real gases have real volume. Also, ideal gases are considered elastic, having no attractive and repulsive forces with no energy transfer during collisions. Real gases actually collide and are non-elastic. Note that gases approach ideal behavior as their temperature increases and their pressure decreases.

Example Question #1 : Real Gases And Ideal Gases

Which of the following assumptions is not made by the ideal gas law?

Possible Answers:

The size of the molecules is much smaller than the container

The molecules obey Newton's laws of motion at all times

The molecules move randomly

The van der Waals forces are negligible

The intermolecular interactions follow the Coulomb model of electric repulsion

Correct answer:

The intermolecular interactions follow the Coulomb model of electric repulsion

Explanation:

Under the ideal gas law, we assume that the interactions between the molecules are very brief and that the forces involved are negligible. The assumption that the molecules obey Coulomb's law when interacting with each other is not necessary; rather, an ideal gas must disregard Coulomb's law.

The ideal gas law assumes only Newtonian mechanics, disregarding any intermolecular or electromagnetic forces.

Example Question #26 : Gases

Consider a real gas with a constant amount and a constant pressure. It has a temperature of  and a volume of . If you double the temperature, what will happen to the volume?

Possible Answers:

The volume will become 

The volume will become less than 

The volume will become greater than 

The volume will become 

Correct answer:

The volume will become less than 

Explanation:

This question can be solved using either Charles's law or the ideal gas law (converted into the combined gas law).

Charles's Law:

Ideal Gas Law: 

The question states that the pressure and moles  are held constant; therefore, the volume and temperature are directly proportional. If the question were asking about an ideal gas, the volume would double when you double the temperature

The volume would double for an ideal gas; however, the question is asking about a real gas. To find the correct relationship between volume and temperature we need to look at the equation for real gas volume. Remember that the volume we are concerned with is the volume of the free space in the container, given by the container volume minus the volume of the gas particles. The equation for real gas volume accounts for the volume of the container and the volume of the gas particles. For a real gas, the volume is given as follows:

In this equation,  is the number of moles of gas particles and  is the bigness coefficient. This equation implies that the volume of free space for a real gas is always less than the volume for an ideal gas; therefore, doubling the temperature will produce a volume that is less than the predicted volume for an ideal gas. Our answer, then, must be less than double the initial volume.

Note that for an ideal gas the bigness coefficient, , would be zero and the volume of free space  would be equal to the volume of the container . This occurs because the volume of the gas particles is negligible for an ideal gas.

Example Question #61 : Fluids And Gases

Which of the following is relevant for real gases, but irrelevant for ideal gases? 

I. Volume of gas particles

II. Intermolecular forces between gas particles

III. Volume of container

Possible Answers:

I only

I and III

III only

I and II

Correct answer:

I and II

Explanation:

There are two main assumptions for an ideal gas (and a few smaller assumptions). First, the gas particles of the ideal gas must have no molecular volume. Second, the gas particles must exert no intermolecular forces on each other; therefore, forces such hydrogen bonding, dipole-dipole interactions, and London dispersion forces are irrelevant in ideal gases. Other small assumptions of ideal gases include random particle motion (no currents), lack of intermolecular interaction with the container walls, and completely elastic collisions (a corollary of zero intermolecular forces).

For real gases, however, these assumptions are invalid. This means that the real gas particles have molecular volume and exert intermolecular forces on each other. 

Recall that the volume in the ideal gas law is the volume of the free space available inside the container. For ideal gases, the free space volume is equal to the volume of the container because the gas particles take up no volume; however, for real gases, the free space volume is the volume of the container minus the volume of the gas particles. Though the exact values of free space volume will differ, the volume of the container is important for both real and ideal gases.

Example Question #1 : Quantum Chemistry

Which of the following sets of quantum numbers () could correspond to a  orbital?

Possible Answers:

Correct answer:

Explanation:

Since we're looking at the  orbital, we know . The range of possible values for  is 0 to . Possible values for  range  to . Therefore, among the answer choices,  is the only possible combination of quantum numbers corresponding to an  orbital. 

Example Question #1 : Quantum Numbers

Which of the following sets of quantum numbers could correspond to a  orbital?

Possible Answers:

Correct answer:

Explanation:

Since we're looking at the  orbital, we know . The range of possible values for  is 0 to . Possible values for  range  to . Therefore, among the answer choices,

is the only possible combination of quantum numbers corresponding to a  orbital.

Example Question #1111 : Mcat Physical Sciences

A scientist is studying a reaction, and places the reactants in a beaker at room temperature. The reaction progresses, and she analyzes the products via NMR. Based on the NMR readout, she determines the reaction proceeds as follows:

In an attempt to better understand the reaction process, she varies the concentrations of the reactants and studies how the rate of the reaction changes. The table below shows the reaction concentrations as she makes modifications in three experimental trials.

Reaction rates depend on reaction concentrations, as well as the reaction constant . The Arrhenius equation is used to calculate this constant. Which of the following factors is used in the Arrhenius equation?

I. Temperature

II. Activation energy

III. Reactant concentrations

Possible Answers:

I and II

II and III

I, II, and III

III only

I only

Correct answer:

I and II

Explanation:

If you thought all of these factors are used in the Arrhenius equation, consider that the rate constant is independent of the reactant concentrations. The concentration effect is considered in the rate law by multiplying the concentrations by the constant, but the concentrations themselves are independent of the actual Arrhenius calculation.

The calculation shows that . In this equation, is the activation energy and is the temperature.

Learning Tools by Varsity Tutors