A monatomic ideal gas at 27° C undergoes a constant pressure process from A to B and a constant volume process from B to C. Find the total work done during these two processes. (Let V1 = 4.70 L and and P1 = 2.50 atm.)
http://img690.imageshack.us/i/15p005alt.gif/
http://img690.imageshack.us/i/15p005alt.gif/
-
The general idea is that the work done by a monotonic gas is defined by the change in volume as well as pressure. In the first process, there is a constant pressure but volume is doubles. At a constant pressure, the work done is equal to the pressure times the change in volume, or w=P*delta(V). Using the numbers:
deltaV = final volume - initial volume = 2*4.7 - 4.7 = 4.7, and P=2.5*2=5
thus, W = 4.7*5 = 23.5 liter*atms
Now the second process has no change in volume. In principle, for an isovolumetric process, there is no work done. Thus, the answer is simple the one stated above., 23.5 liter*atms = 2380 J
Idk how interested ur into calculus or math at all, but the work does by a process can be determined by the integral for the PV graph, or simply, the area between the PV graph and the x axis.
deltaV = final volume - initial volume = 2*4.7 - 4.7 = 4.7, and P=2.5*2=5
thus, W = 4.7*5 = 23.5 liter*atms
Now the second process has no change in volume. In principle, for an isovolumetric process, there is no work done. Thus, the answer is simple the one stated above., 23.5 liter*atms = 2380 J
Idk how interested ur into calculus or math at all, but the work does by a process can be determined by the integral for the PV graph, or simply, the area between the PV graph and the x axis.