#### Question

A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Figure

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F

#### Solution 1

Total work done by the gas from D to E to F = Area of ΔDEF

Area of ΔDEF =`1/2DExxEF`

Where,

DF = Change in pressure

= 600 N/m^{2} – 300 N/m^{2}

= 300 N/m^{2}

FE = Change in volume

= 5.0 m^{3} – 2.0 m^{3}

= 3.0 m^{3}

Area of ΔDEF =`1/2xx300xx3` = 450 J

Therefore, the total work done by the gas from D to E to F is 450 J.

#### Solution 2

As is clear figure

Change in pressure, `triangle P = EF = 5.0 - 2.0 = 3.0 atm = 3.0 xx 10^5 Nm^(-2)`

Change in volume. triangle V = DF = 600 - 300 = 300 cc

= `300 xx 10^(-6) m^3`

Work done by the gas from D to E to F = area of `triangle DEF`

`W =1/2 xx DF xx EF`

`= 1/2 xx (300 xx 10^(-6)) xx (3.0 xx 10^5) = 45 J`