Question

The properties of a closed system change following the relation between pressure and

volume as pv = 3.0 where p is in bar V is in m^3. Calculate the work done when the pressure

increases from 1.5 bar to 7.5 bar.
can anyone solve this...​

Answer 1

Good night............................

Answer 2

Explanation:

Given The properties of a closed system change following the relation between pressure and volume as pv = 3.0 where p is in bar V is in m^3. Calculate the work done when the pressure increases from 1.5 bar to 7.5 bar

• Given P = 1.5 bar to 7.5 bar = 1,50,000 Pascal to 7,50,000 Pascal
• So we have P x V = constant
• Now P x V = 3
• Or P = 3/V
• Now when P = 1,50,000 pa
• P = 3/V
• 1,50,000 = 3/V
• V = 3/1,50,000
• V = 2 x 10^-5 m^3
• Now when P = 7,50,000 pa
• P = 3/V
• 7,50,000 = 3/V
• V = 3/7,50,000
• V = 0.4 x 10^-5 m^3
• So work done is given by
•   W = ∫V1 to V2 P dV
•       = ∫(2 to 0.5)10^-5 3/V dV
•        = 3 log (0.4 / 2)
•       = - 3 x 0.69
•        = - 2.07 J
• Negative sign shows work done on the system  

Reference link will be

https://brainly.in/question/2174883