Question

. 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 the volume (p = a + bV). The internal energy
of the fluid is given by the following equation
U = 42 + 3.6 pV
where U is in kJ, p in kPa, and V in cubic metre. If the fluid changes from an initial state of
190 kPa, 0.035 m3 to a final state of 420 kPa, 0.07 m3, with no work other than that done on the
piston, find the direction and magnitude of the work and heat transfer.

Answer 1

$p=a+bV \\ p_1 =190 \;kPa \\ V_1= 0.035 \;m^3 \\ p_2= 420 \;kPa \\ V_2=0.07 \;m^3 \\ p_1= a+bV_1 \\ p_2= a+bV_2 \\ p_2-p_1=b(V_2-V_1) \\ b = \frac{p_2-p_1}{V_2-V_1} \\ p_1= a+bV_1 \\ p_1 = a+ (\frac{p_2-p_1}{V_2-V_1})V_1 \\ a= p_1 -(\frac{p_2-p_1}{V_2-V_1})V_1 \\ = \frac{p_1V_2 -p_1V_1 -p_2V_1+p_1V_1}{V_2-V_1} \\ = \frac{p_1V_2-p_2V_1}{V_2-V_1} \\ b = \frac{420-190}{0.07-0.035} = 6571.43 \;kPa/m^3 \\ a=\frac{190 \times 0.07 + 420 \times 0.035}{0.07-0.035} = -40 \;kPa \\ p= a+bV \\ p= -40+6571.43V$

Work transfer (W) $= \int^{V_2}_{V_1}pdV$

$= \int^{0.07}_{0.035} (-40 +6571.43V)dV \\ = (-40V + \frac{6571.43}{2}V^2)^{0.07}_{0.035} \\ = -40(0.07 -0.035) + \frac{6571.43}{2}(0.07^2 -0.035^2) \\ = -1.4 + 12.075 \\ = 10.675 \;kJ$

Since work is positive, work is done by the fluid.

$U = 42 + 3.6 pV \\ U_1=42+3.6p_1V_1 \\ U_2=42+3.6p_2V_2 \\ ΔU=U_2-U_1 \\ = 3.6(p_2V_2- p_1V_1) \\ = 3.6(420 \times 0.07 -190 \times 0.035) \\ = 81.9 \;kJ$

Heat transfer Q = ΔU+W

$= 81.9+10.675 \\ = 92.575 \;kJ$

Since heat transfer is positive, heat is transferred to the system.