# Calculating work in a one-step expansion/compression

• Footballer010
In summary, the conversation discusses a problem with a thermodynamics process involving an ideal gas. The speaker suggests using the ideal gas equation and Boyle's law to calculate the work for each step of the process. The middle step can be ignored if only the net transition needs to be calculated.
Footballer010
Hello. I'm having problems with one thermodynamics. I've used several equations but I'm still unable to get the right answer. A push in the right direction would be helpful. Thanks.

Consider a process involving 1 mole of an ideal gas that takes place by the following pathway, at a constant temperature of 25oC:
P1 = 3 atm → P2 = 0.75 atm → P3 = 3 atm
Both steps occur irreversibly.

work=?

W=P[delta]V

Boyle's Law: P1V1=P2V2

PV=nRTSo, use the ideal gas equation (PV=nRT) to solve for the initial volume of gas present. Then you can use Boyle's law to calculate the new volume after the change has occurred (the new volume is V2, the initial pressure is P1, etc) you can then multiply the new pressure by the CHANGE in volume (ie, the difference between V1 and V2) to get the work. Repeat process for the next step of the transfer. If you only need to calulate work for the net transition, completely ignore the middle step.

Hello,

Calculating work in a one-step expansion/compression can be done using the equation W = -PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume. In this case, since the temperature is constant, the change in volume can be calculated using the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

To calculate the work done in the first step, we can use the initial pressure (P1 = 3 atm) and the final pressure (P2 = 0.75 atm) to find the change in pressure (ΔP = P2 - P1 = -2.25 atm). We can then use this value to find the change in volume (ΔV = nRT/ΔP = (1 mol)(0.0821 L atm/mol K)(298 K)/(-2.25 atm) = -108.9 L).

To calculate the work done in the second step, we can use the final pressure (P3 = 3 atm) and the change in volume from the first step (ΔV = -108.9 L) to find the work done (W = -PΔV = (-3 atm)(-108.9 L) = 326.7 L atm).

Therefore, the total work done in the process is 326.7 L atm. Please note that since both steps occur irreversibly, the work done in this process is not the maximum possible work. I hope this helps to guide you in the right direction. Best of luck in your thermodynamics calculations.

## What is work in a one-step expansion/compression?

Work is defined as the amount of energy transferred to or from a system due to a force acting on it. In the case of a one-step expansion or compression, work is done on the system by changing its volume.

## How is work calculated in a one-step expansion/compression?

The work done in a one-step expansion or compression can be calculated using the formula W = PΔV, where W is work, P is the pressure, and ΔV is the change in volume.

## What units are used to measure work in a one-step expansion/compression?

Work is typically measured in joules (J) or newton-meters (N∙m) in the SI system. In some cases, work may also be measured in other units such as calories (cal) or foot-pounds (ft∙lb).

## Can work be negative in a one-step expansion/compression?

Yes, work can be negative in a one-step expansion or compression if the system is doing work on its surroundings. This means that the system is losing energy and the surroundings are gaining energy.

## What factors affect the amount of work in a one-step expansion/compression?

The amount of work in a one-step expansion or compression is affected by the pressure applied to the system, the change in volume, and the direction of the work (whether it is done on or by the system).

• Introductory Physics Homework Help
Replies
2
Views
808
• Chemistry
Replies
3
Views
4K
• Thermodynamics
Replies
4
Views
871
• Introductory Physics Homework Help
Replies
5
Views
3K
• Thermodynamics
Replies
8
Views
670
• Thermodynamics
Replies
22
Views
2K
• Thermodynamics
Replies
3
Views
904
• Thermodynamics
Replies
45
Views
2K
• Thermodynamics
Replies
3
Views
1K
• Other Physics Topics
Replies
3
Views
1K