# Solve Ideal Gas Problem Homework: Compute Work Done by Air

• jeff.berhow
In summary, a problem involving an initial volume and gauge pressure of air being expanded isothermally and then cooled at constant pressure until it reaches its initial volume is being solved. The work done by the air is calculated using the formula Wisothermal + Wisobaric = Wtotal. However, the second pressure given in the problem should be treated as actual pressure instead of gauge pressure, resulting in the correct answer of 5.6 kJ. This highlights the importance of reading and understanding the given information in a problem.
jeff.berhow

## Homework Statement

Air that initially occupies 0.140m3 (V1) at a gauge pressure of 103.0 kPa (p1) is expanded isothermally to a pressure of 101.3 kPa (p2) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air.

## Homework Equations

Wisothermal = nRTln(V2/V1)
Wisobaric = p2ΔV = p2(V1 - V2)
Wisothermal + Wisobaric = Wtotal

## The Attempt at a Solution

We can find V2 easily enough by multiplying p1 and V1 and dividing by p2 because both p1V1 and p2V2 equal nRT. I can now fill in most of the formula:

nRTln(V2/V1) + p2(V1 - V2) = Wtotal

Here, I thought I could then just back substitute nRT for either p1V1 or p2V2 and I'd be golden but the answer I'm getting is different from the back of the book (5.6 kJ). I'm getting .001 kJ with plugging these in:

Wtotal = (103kPa)(0.140m3)ln(0.140m3/.141m3) + (101.3 kPa)(.140m3 - .141m3)

Unfortunately, if I use the actual pressure, I still don't get the desired answer:

Wtotal = (204kPa)(0.140m3)ln(0.140m3/.141m3) + (202.3 kPa)(.140m3 - .141m3)

What am I fundamentally missing here? Thanks in advance.

Doc Al said:
They may be pulling a sneaky one on you. Only the first pressure given is specified as gauge pressure. Treat the second pressure as actual pressure. (That would make this a much more realistic problem!)

Wtotal = (204kPa)(0.140m3)ln(0.140m3/.141m3) + (101.3 kPa)(.140m3 - .141m3) = 5.6 kPa

Last edited:
For one thing, how does gauge pressure relate to absolute pressure?

Gauge pressure is the difference between the atmospheric pressure and the actual pressure in the object.

jeff.berhow said:
Gauge pressure is the difference between the atmospheric pressure and the actual pressure in the object.
Right. But in your calculation, you used the gauge pressure. (At least that's what you wrote.)

Doc Al said:
Right. But in your calculation, you used the gauge pressure. (At least that's what you wrote.)

You're absolutely right, Doc Al. I've updated my original post to reflect that. Thanks! :)

jeff.berhow said:
Air that initially occupies 0.140m3 (V1) at a gauge pressure of 103.0 kPa (p1) is expanded isothermally to a pressure of 101.3 kPa (p2) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air.
They may be pulling a sneaky one on you. Only the first pressure given is specified as gauge pressure. Treat the second pressure as actual pressure. (That would make this a much more realistic problem!)

That was it! Man, it feels like this is a reading comprehension problem more than an actual physics problem. I was really worried that I wasn't understanding something very fundamental about pressures, temperatures and volumes.

Thanks Doc Al!

## 1. What is an ideal gas problem?

An ideal gas problem is a type of calculation that involves using the Ideal Gas Law equation, PV = nRT, to solve for one of the variables (pressure, volume, number of moles, or temperature) when the others are given. This equation describes the relationship between the physical properties of a gas at a constant temperature.

## 2. How do you compute work done by air in an ideal gas problem?

To compute the work done by air in an ideal gas problem, you can use the formula W = -PΔV, where P is the pressure and ΔV is the change in volume. This formula is derived from the area under the pressure-volume curve on a P-V diagram. It represents the energy transferred to or from the gas due to a change in its volume.

## 3. What are the units for work done by air in an ideal gas problem?

The units for work done by air in an ideal gas problem are Joules (J). This is a unit of energy and is equivalent to 1 kg*m^2/s^2.

## 4. How does the temperature affect the work done by air in an ideal gas problem?

In an ideal gas problem, the work done by air is directly proportional to the change in temperature. This means that as the temperature increases, the work done by air also increases. This relationship is described by the Ideal Gas Law equation, where temperature is directly related to the pressure and volume of the gas.

## 5. Can the work done by air in an ideal gas problem be negative?

Yes, the work done by air in an ideal gas problem can be negative. This occurs when the gas is compressed and the volume decreases. In this case, the change in volume (ΔV) is negative, resulting in a negative value for work done. This means that the gas is losing energy and work is being done on the gas by an external force.

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