Cisco
CompTIA
ISC
EC-Council
EXIN
Fortinet
ITIL
Juniper
Microsoft
VMware
Avaya
SAP
Google
NVIDIA
Salesforce
Amazon
Palo Alto Networks
ISC2
Splunk
ServiceNow
All Vendors
  2. Home
  3. Exams
  4. Test Prep
  5. MCAT Test

Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 25 )

Updated On: 30-Jan-2026
Page 25 of 164
PDF with 811 Questions

The equation of state of an ideal gas is given by the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.

If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision.
The average kinetic energy (E) of a gas particle is given by:

Equation 1
where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles:

where N is the number of gas particles; this is different from the average speed. For an ideal gas, the kinetic energy of all the particles is:

Equation 2
where n is the number of moles of gas. Combining these equations gives:


Equation 3
where M is the molar mass of the gas particles.
The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct.

Equation 4
In this equation, s is the atomic diameter (typically on the order of 10-8), k is the Boltzmann constant, and P is the pressure.
In addition to colliding with one another, gas particles also collide with the walls of their container. If the container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

Equation 5
Where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P1.
What is the relative rate of effusion for a mixture of two noble gases, GA and GB, which escape through the same pinhole?

  1. 1



Answer(s): C

Explanation:

If the ratio of the rates of effusion for the two gases is taken, and all factors that are the same for the two gases are canceled out, then one ends up simply with the inverse of the square root of the masses, as in choice C.
Since the gases A and B constitute a mixture, they are in the same container, at the same pressure and effusing through the same pinhole. Therefore, the pinhole area for A equals the pinhole area of B, and the answer given in choice B can be condensed to the answer given in choice C. Similarly, the pressure of the gas
in the container is the same whether you are considering the gas A component or the gas B component, since they are in a mixture characterized by one pressure. Thus, the answer given in choice D also reduces to that in choice C. Choice A, which states that the ratio is 1, would only be correct if the two gases being considered had the same molecular weight. Although it is possible for this to be true, we have no indication of whether or not it is. Since the question states that they are two noble gases, it is safer to assume that the two gases have different molecular weights.



The equation of state of an ideal gas is given by the ideal gas law:
= nRT
PV
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.

If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision.
The average kinetic energy (E) of a gas particle is given by:

Equation 1
where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles:

where N is the number of gas particles; this is different from the average speed. For an ideal gas, the kinetic energy of all the particles is:

Equation 2
where n is the number of moles of gas. Combining these equations gives:

Equation 3
where M is the molar mass of the gas particles.

The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct.

Equation 4
In this equation, s is the atomic diameter (typically on the order of 10-8), k is the Boltzmann constant, and P is the pressure.
In addition to colliding with one another, gas particles also collide with the walls of their container. If the container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

Equation 5
Where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P1.
The average kinetic energy of an ideal gas can be directly related to the:

  1. rms speed.
  2. temperature.
  3. Boltzmann constant.
  4. universal gas constant.

Answer(s): B

Explanation:

Equation 2 gives the formula for the average kinetic energy of an ideal gas. To answer this question, all you have to do is figure out which factor varies directly with the average kinetic energy. The rms speed, u, has been canceled out of this equation, so choice A is wrong. The Boltzmann constant had universal gas constant, choices C and D, are the same for every value of the average kinetic energy, so they are also wrong. That leaves only temperature, choice B, the correct answer. All that is needed to obtain the correct answer is Equation 2 and an understanding of the various variables in the equation. However, choice B may have been your instinctive choice since a study of gases always mentions that the temperature of a gas is a measure of the average kinetic energy of that gas. Either way, the correct answer is B.



The equation of state of an ideal gas is given by the ideal gas law:
= nRT
PV
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.


If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision.
The average kinetic energy (E) of a gas particle is given by:

Equation 1
where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles:

where N is the number of gas particles; this is different from the average speed. For an ideal gas, the kinetic energy of all the particles is:

Equation 2
where n is the number of moles of gas. Combining these equations gives:

Equation 3
where M is the molar mass of the gas particles.
The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct.

Equation 4
In this equation, s is the atomic diameter (typically on the order of 10-8), k is the Boltzmann constant, and P is the pressure.

In addition to colliding with one another, gas particles also collide with the walls of their container. If the container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

Equation 5
Where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P1.
Which of the following will have the smallest root mean square speed at 298K?

  1. Cl2(g)
  2. O2(g)
  3. CO2(g)
  4. N2(g)

Answer(s): A

Explanation:

Equation 3 states that the root mean square speed, u, is equal to

Therefore, u is proportional to

Since the temperature is held constant, we must look at the molar mass. If u is proportional to

the molecule with the largest molar mass will have the smallest root mean square speed. Chlorine, choice A, has the largest molar mass (71 g/mol) and is the correct response. Oxygen, carbon dioxide, and nitrogen have molar masses of 32 g/mol, 44 g/mol, and 28 g/mol, respectively. When compared with chlorine, we can see that all of these molecules will have a larger root mean square speed.



The periodic beating of the heart is controlled by electrical impulses that originate within the cardiac muscle itself. These pulses travel to the sinoatrial node and from there to the atria and the ventricles, causing the cardiac muscles to contract. If a current of a few hundred milliamperes passes through the heart, it will interfere with this natural system, and may cause the heart to beat erratically. This condition is known as ventricular fibrillation, and is life-threatening. If, however, a larger current of about 5 to 6 amps is passed through the heart, a sustained ventricular contraction will occur. The cardiac muscle cannot relax, and the heart stops beating. If at this point the muscle is allowed to relax, a regular heartbeat will usually resume.

The large current required to stop the heart is supplied by a device known as a defibrillator. A schematic diagram of a defibrillator is shown below. This device is essentially a "heavy-duty" capacitor capable of storing large amounts of energy. To charge the capacitor quickly (in 1 to 3 seconds), a large DC voltage must be applied to the plates of the capacitor. This is achieved using a step-up transformer, which creates an output voltage that is much larger than the input voltage. The transformer used in this defibrillator has a step-up ratio of 1:50.

The AC voltage that is obtained from the transformer must then be converted to DC voltage in order to charge the capacitor. This is accomplished using a diode, which allows current flow in one direction only. Once the capacitor is fully charged, the charge remains stored until the switch is moved to position B and the plates are placed on the patient's chest. To cut down the resistance between the patient's body and the defibrillator, the electrodes are covered with a wetting gel before use. Care must be taken to insure that the patient is not in electrical contact with the ground while the defibrillator is in use.
If the defibrillator has a capacitance of 10 F, how much charge will build up on the two plates?

  1. 0.08 coulombs
  2. 0.8 coulombs

Answer(s): A

Explanation:

The correct answer choice here is 0.08 coulombs. To do this question, you must remember the equation Q = , where Q is the charge stored, C is the capacitance, and V is the voltage across the plates. We are given CV
that the capacitance is 10 micro farads, but we do not have a value for the voltage V, since the input voltage to the transformer is stepped-up 50 times. From the diagram we can see that the input voltage is 160 volts. So the output voltage is 160 times 50, or 8,000 volts. Substituting the voltage and the capacitance into our equation, gives a value for the charge of 8,000 times 10 × 10-6, or 0.08 coulombs, choice A.



The periodic beating of the heart is controlled by electrical impulses that originate within the cardiac muscle itself. These pulses travel to the sinoatrial node and from there to the atria and the ventricles, causing the cardiac muscles to contract. If a current of a few hundred milliamperes passes through the heart, it will interfere with this natural system, and may cause the heart to beat erratically. This condition is known as ventricular fibrillation, and is life-threatening. If, however, a larger current of about 5 to 6 amps is passed through the heart, a sustained ventricular contraction will occur. The cardiac muscle cannot relax, and the heart stops beating. If at this point the muscle is allowed to relax, a regular heartbeat will usually resume.
The large current required to stop the heart is supplied by a device known as a defibrillator. A schematic diagram of a defibrillator is shown below. This device is essentially a "heavy-duty" capacitor capable of storing large amounts of energy. To charge the capacitor quickly (in 1 to 3 seconds), a large DC voltage must be applied to the plates of the capacitor. This is achieved using a step-up transformer, which creates an output voltage that is much larger than the input voltage. The transformer used in this defibrillator has a step-up ratio

of 1:50.

The AC voltage that is obtained from the transformer must then be converted to DC voltage in order to charge the capacitor. This is accomplished using a diode, which allows current flow in one direction only. Once the capacitor is fully charged, the charge remains stored until the switch is moved to position B and the plates are placed on the patient's chest. To cut down the resistance between the patient's body and the defibrillator, the electrodes are covered with a wetting gel before use. Care must be taken to insure that the patient is not in electrical contact with the ground while the defibrillator is in use.
The resistance between the two electrodes when placed apart on the patient's chest is 1,000 when wetting gel is used. What is the initial current through the patient's heart, assuming that all the current takes this path?

  1. 0.16 A
  2. 4 A
  3. 6.25 A
  4. 8 A

Answer(s): D

Explanation:

This question asks you to calculate the initial current through the patient, and this can be done by using Ohm's law which states that the current is equal to the voltage divided by the resistance. We know that the resistance across the two electrodes is 1000 ohms when a wetting gel is applied, but as in the previous question we do not have a value for the initial voltage. This be calculated as well. We know that the transformer increases the input voltage by a factor of 50, and since the input voltage is 160 volts, the output voltage will be 8000 volts. This voltage charges the capacitor only, so the voltage across the capacitor when fully charged will be 8000 volts. By substituting the values for the resistance and the voltage into Ohm's law, we find that the initial current flowing through the patient is 8000 divided by 1000, or 8 amps which is choice D.



Viewing page 25 of 164
Viewing questions 121 - 125 out of 811 questions



Post your Comments and Discuss Test Prep MCAT Test exam prep with other Community members:

Join the MCAT Test Discussion