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## Q. 2.37

A fluid is contained in a system behind a spring loaded, frictionless piston so that the pressure in the fluid is given by:

p = a + bV

The internal energy of the fluid is given by:

U = 45 + 3.5 pV kJ

Where p is in kPa and V in m³. The fluid changes from an initial state of 200 kPa, 0.04 m³ to a final state of 450 kPa, 0.08 m³. Find the amount of work done and heat transfer.

## Verified Solution

\begin{aligned}d U &=U_2-U_1 \\&=\left(45+3.5 p_2 V_2\right)-\left(45+3.5 p_1 V_1\right) \\&=3.5(450 \times 0.08-200 \times 0.04)=98 \mathrm{~kJ}\end{aligned}

Now                   p = a + bV
200 = a + 0.04 b
450 = a + 0.08 b

\begin{aligned}b &=\frac{250}{0.04}=6250, \quad a=200-0.04 \times 6250=-50 \\p &=-50+6250 \mathrm{~V} \\W_{1-2} &=\int_1^2 p d V=\int_{0.04}^{0.08}(-50+6250 \mathrm{~V}) d V \\&=\left|-50 \mathrm{~V}+3125 \mathrm{~V}^2\right|_{0.04}^{0.08} \\&=-50(0.08-0.04)+3125\left(0.08^2-0.04^2\right)=-2+15=13 \mathrm{~kJ} \\\mathrm{Q}_{1-2} &=d U+W_{1-2}=98+13=111 \mathrm{~kJ}\end{aligned}