Calculating work in a one-step expansion/compression

[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=?

 

[wil3]

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.

 

[quicknote]

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.