Question

Determine the work (as a factor of x10⁵ Joules) from the pV diagram shown. The gas is taken from point A = (2 m³, 3x10⁵ N/m²) to B = (4 m³, 6x10⁵ N/m² ) and then to C = (6 m³, 6x10⁵ N/m²).

	a.	16
	b.	21
	c.	18
	d.	32

Answer 1

PC x105 N/m?) Points A, B and a are shown in pu diagram. B C 6 5 4 3 A G 2 IF E D 2 3 4 5 6 + + v (m3) work done = Area under the curve ABC = Area of triangle ABG + Area of rectangle AFEG of rectangle + Area le BC DE Area of AG 2 m3 = 2m3 Then, = 13x10 New Area triangle ABG = Į (BGxAG) 6x105 N/m2 - 3x105 N/m²= 3x10 N/m2 4 m3 Area of triangle ABG = Į (3x105 N/m?) (2m3) of rectangle de AFEG= AF XFE AF = 3x10 N/m² - ON/m2 = 3x105 N/m2 FE = 4m²_zm3 am3 Area of rectangle AF EG = 3x10$ Nym?) (2013) = 16x10 Nm Area of rectangle BCDE = BC XBE 4m3 = 2m 6x105 N/m² - ON/m² = 6810S W/m² Then, Area of rectangle B C D E = (2m²) (6X10SN/m²) This work done = 3x105 Nm + 6X105 Nm +12X10Nm =(21x10s Thus, option 6 is correct. Then, Now, Вс. 6 m3 BE - (12x105 Nm