Discuss the below:
Simulation Investigating Isobaric, Isochoric, & Isothermal Processes
Perform a simulation investigating three special processes - isobaric (constant pressure), isochoric (constant volume), and isothermal (constant temperature) - that can be derived from the ideal gas law. The simulation is found at Background Info 1. The simulation allows you to vary the pressure for an isobaric process, the volume for an isochoric process, and the temperature for an isothermal process. The simulation shows how the system goes from one equilibrium state to another and the work that is done on or by the system, and how heat flows into or out of the system.
The variables are:
V1 = initial volume, dm3 (dm = decimeter = 10 cm)
V2 = final volume, dm3
P1 = initial pressure, kPa (absolute; that is, measured against vacuum)
P2 = final pressure, kPa
T1 = initial temperature, K
T2 = final temperature, K
Heat flow: into system (red arrow points right) out of system (red arrow points left)
Work done: on system (blue arrow points up) by system (blue arrow points down)
Experiment 1:
Isobaric (constant pressure) Process
P1 = P2 = 100 kPa
Experiment 2:
Isochoric (constant volume) process
V1 = V2 = 1.00 dm3
Following a procedure similar to the one in Experiment 1, show how these results illustrate the Ideal Gas Law for the special case of constant volume.
Experiment 3:
Isothermal (constant temperature) p
Following a procedure similar to the one in Experiment 1, show how these results illustrate the Ideal Gas Law for the special case of constant temperature.
The isothermal case deserves a special mention, because at first glance it seems to be just plain wrong. Everybody knows that when you squeeze a quantity of gas, the volume goes down and the pressure goes up; but the temperature ALSO goes up. Ask any diesel mechanic! So what's going on here?