A fluid is contained in a cylinder by a spring loaded, frictionless piston so that the pressure in the fluid is a linear function of volume (p = a +bV). The internal energy of the fluid in kJ is given by the expression U= 32 +3 pV where p is in kPa and V is in mo. The initial and final pressures are 150 kPa and 350 kPa and the corresponding volumes are 0.02 m^3 and 0.05 m^3Make calculations for the direction and magnitude of work and heat interactions.

A fluid is contained in a cylinder by a spring loaded, frictionless pi

1.	A fluid is contained in a cylinder by a spring loaded, frictionless piston so that the pressure in the fluid is a linear function of volume (p = a +bV). The internal energy of the fluid in kJ is given by the expression U= 32 +3 pV where p is in kPa and V is in mo. The initial and final pressures are 150 kPa and 350 kPa and the corresponding volumes are 0.02 m^3 and 0.05 m^3Make calculations for the direction and magnitude of work and heat interactions.
Answer» Solution :Relation between pressure and volume, p= a +bV. The values of constant a and b can be determined form the values of pressure and temperature at the initial and final state POINTS. `150 = a +0.02 b and 350 = a +0.05 b` From these expressions : a = 16.67 and b = 6666.67 The LAW of expansion BECOMES : p= 16.67 +6666.67 V a) Work involved during the process `W_(12)= int_(1)^(2)p dV= int_(1) ^(2)( 16.67 + 6666.670) dv` `= 16.67(V_2 -V_1)+(6666.67 )/(2)(V_(2)^(2) - V_(1)^(2))` `= 16.67 (V_2 -V_1) +(6666.67 )/(2)( 0.05^2- 0.02^2) = 7.5 J` SINCE `W_(12)`is positive, the work has been done by the system, the magnitude being 7.5 kJ. (b) Change in internal energy of the fluid during the process `U_2 -U_1-(32-3p_2 - V_2)-(32 - 3p_1 . V_1)` Applying non-flow energy equation , `delta Q =delta W + dU.` the heat interaction is `Q_(12)= 7.5 + 43.50 = 50.50 kJ` since ` Q_2`is positivetheheatflowsintothe system, themagnitudebeing50.5 KJ.