College Physics : Thermodynamics

Study concepts, example questions & explanations for College Physics

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Example Questions

Example Question #1 : Laws Of Thermodynamics And General Concepts

A  steel sword is cooled from  to  by dipping it in  of water that is at

If the specific heat of  the steel is  and the specific heat of water is , what will be the temperature of the water after the steel is cooled?

Possible Answers:

Correct answer:

Explanation:

This is a simple pairing of two  equations.

We will make the left side for the steel and the right side for water. Thus:

Solving for  of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is 

Example Question #1 : Thermodynamics

A steel beam is  long at a temperature of . On a hot day, the temperature reaches . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is

Possible Answers:

Correct answer:

Explanation:

We need to use the equation for thermal expansion in order to solve this problem:

We are given:

for the thermal expansion constant

 for the initial length of the steel.

We need to calculate which is the the difference of the two temperatures.

Now we have enough information to solve for the change in the length of steel.

Example Question #1 : Thermodynamics

What is the change in entropy for  of ice initially at  that melts to water and warms to the temperature of the atmosphere at ?

Latent heat of fusion for water: 

Specific heat of water: 

Specific heat of ice:  

Possible Answers:

Correct answer:

Explanation:
  • We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

Where  is the change in entropy,  is the change in heat energy, and  is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for  for when ice melts: . Our entropy equation now becomes:

Now we plug in our known values into the equation:

Mass 

Latent heat of fusion 

Temperature 

And we get the change in entropy for the ice melting at a constant temperature, which is 

  • Now we need to calculate the change in entropy for the liquid water warming from  to .

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

Next we substitute in an equation for when there is a temperature change, but not a change in state. That equation is: so our entropy equation now becomes:

next we integrate this equation to get:

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water 

Initial temperature of water 

Specific heat of water 

Mass of water 

Now we get the change in entropy for the water warming to be 

We add our two entropy values together to find the total change in entropy to be:

Example Question #1 : Ideal Gases

If an ideal gas has its temperature doubled while its volume is cut in half, what happens to its pressure?

Possible Answers:

The pressure increases by a factor of 

The pressure increases by a factor of 

There is no change in the pressure

The pressure decreases by a factor of 

Correct answer:

The pressure increases by a factor of 

Explanation:

In this question, we're told that an ideal gas undergoes a change in its temperature and volume, and we're asked to determine how its pressure changes.

First, we'll need to recall the ideal gas equation.

Then, we can rearrange to isolate the term for pressure.

From this expression, we can see that if we double the volume, the pressure will double. Also, cutting the volume in half would double the pressure as well. Therefore, doing both of these would cause the pressure to increase by a factor of .

Example Question #1 : College Physics

What volume does  of oxygen gas occupy at STP?

Possible Answers:

Correct answer:

Explanation:

 The first step in this question is to convert the mass of oxygen given into moles.

 

Next, it's important to recognize that we'll need to use the ideal gas equation. Since we're told that the gas is under STP (standard temperature and pressure), we have everything we need to solve for volume.

Example Question #1 : Isochoric Processes

What is true regarding the work performed by a system whose volume remains constant and undergoes an increase in pressure?

Possible Answers:

The system does work on its surroundings

The work done by the system increases the temperature of the system

The surroundings does work on the system

The system does no work

Correct answer:

The system does no work

Explanation:

To answer this question, it's important to remember how work is defined when dealing with volume and pressure.

The equation for the work done by a gas on its surrounding is as follows.

With this equation in mind, we can see that the only time work is performed is when there is a change in volume. Since we're told in the question stem that the volume remains constant as the pressure is increased, the final volume will be equal to the initial volume. Thus, the  term will be equal to  and no work will be performed.

Example Question #1 : Isothermal Processes

Suppose that a piston contains a gas which begins at a certain pressure and volume. First, the gas in the piston undergoes an isothermal expansion to triple its initial volume. Next, the gas undergoes an isobaric contraction back to its original volume. Finally, the gas undergoes an isochoric increase in pressure until it has returned to its original pressure. Which of the following is a true statement regarding the work done by this gas?

Possible Answers:

Correct answer:

Explanation:

In this question, we're told the process that a gas takes through various changes in conditions. From this information, we're asked to find a true statement regarding the work done by the gas in this process.

At its initial point, we can define the gas as having an initial pressure and volume as .

First, we're told that the gas goes through an isothermal expansion to triple its volume. Recall that an isothermal process is one in which the temperature of the gas doesn't change. Moreover, because we're dealing with an ideal gas, we can see from the ideal gas equation that a constant temperature means that a changing volume must by accompanied by a change in pressure. Since the volume is being increased by a factor of three, we know that the pressure must be decreasing by a factor of three. Thus, right after the isothermal expansion has finished, the gas will have a pressure and volume of .

Next, we're told that the gas undergoes an isobaric contraction back to its original volume. Thus, we know from this that the pressure won't change, but the volume will return to its original value. Consequently, once the isobaric contraction has come to completion, the gas will have a pressure and volume of .

Finally, we're told that the last process is an isochoric increase of the pressure back to its original amount. Isochoric means that the volume doesn't change in this process. Hence, the gas will now be returned to its original pressure and volume of .

If we were to show this on a pressure vs. volume curve, it would look like the following.

 Cyclic gas process

Remember that any pressure-volume work done by a gas is the area under the curve of a pressure vs. volume diagram. For a cyclic process, as the one in this problem, the work done by the gas will be the area within the above image.

For this problem, we don't need to calculate the exact amount of work done. Instead, we can get a rough estimate by first realizing that the structure shown in the above diagram is close to forming a right triangle, but it is slightly smaller. Thus, if we can use the left and bottom sides of the image above to calculate the area that would be enclosed by a right triangle, we can infer that the work done in this problem will be slightly less than the calculated area.

The area for a triangle is the following.

So, for the image shown above, the bottom length is  and the side length is .  With these values, the area of a right triangle can be calculated.

Finally, since this represents the area for a right triangle, we know that the actual work done is less than this. Hence, the correct answer is:

Example Question #1 : Isothermal Processes

An ideal gas with an initial pressure of and an initial volume of expands isothermally to a volume of . What is the total amount of thermal energy transferred to the gas during this process?

Possible Answers:

Correct answer:

Explanation:

We know that with the ideal gas law that:

We also know with isothermal processes, that the work done on a gas is:

Since the number of moles and the temperature is not given to us in the question, we can make a substitution with the ideal gas law to get:

Now we can plug in our known values into the equation to solve for the amount of work energy transferred through the isothermal process.

 

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