Question

Find the work done (in J) in the quasi-static processes shown below. The states are given as (V, p) values for the points in the pV-plane: 1 (5 L, 6 atm), 2 (7 L, 6 atm), 3 (5 L, 8 atm), 4 (7 L, 1 atm), 5 (3 L, 7 atm), 6 (6 L, 8 atm), and 7 (6 L, 1 atm)

Answer 1

The work done by the gas in a reversible or quasi-static process is equal to the area under the PV curve of the process.

One useful formula for this problem is area of a trapezium

The area of trapezium is

$\dpi{120} A=\frac{1}{2}\times width\times (sum\ of\ parallel\ sides)$

And a useful conversion used in the following answer is

$\dpi{120} 1\ L\ atm= 1\times (10^{-3}m^3)\times (1.01\times 10^5Nm^{-2})$

$\dpi{120} 1\ L\ atm= 101\ Nm=101\ J$

(d)

Area under the 1-> 5 process is the area of trapezium in the following figure:

Tatm batim

The area is

$\dpi{120} A_{15}=\frac{1}{2}\times 2L\times (7atm+6atm)=13\ L\ atm$

The gas is being compressed in this process, the work done by the gas is negative

$\dpi{120} W_{15}=-13\ L\ atm$

$\dpi{120} W_{15}=-13\times 101\ J=-1313\ J$

Similarly area under 5-> 3 process is

Batm wapt Na

The work done is

$\dpi{120} W_{53}=\frac{1}{2}\times 2L\times (7atm+8atm)=15\ L\ atm=1515\ J$

Net work done in process 1->5->3 is

$\dpi{120} W_{153}=W_{15}+W_{53}=-1313\ J+1515\ J=202\ J$

(e)

The work done in process 1->6->2 is

$\dpi{120} W_{162}=W_{16}+W_{62}$

1 Batm Glatm Gatm hi E IL V

$\dpi{120} W_{162}=\frac{1}{2}\times 1L\times (6atm+8atm)+\frac{1}{2}\times 1L\times (8atm+6atm)$

$\dpi{120} W_{162}=14\ L\ atm=1414\ J$

(f)

The work done in process 1->7->2 is

$\dpi{120} W_{172}=W_{17}+W_{72}$

PA Gatimi i N 521 ILLIL→

$\dpi{120} W_{172}=\frac{1}{2}\times 1L\times (6atm+1atm)+\frac{1}{2}\times 1L\times (1atm+6atm)$

$\dpi{120} W_{172}=7\ L\ atm=707\ J$