Question

A certain fluid at 10 bar is contained in a cylinder behind a piston, the initial volume
being 0.05 m^3. Calculate the work done by the fluid when it expands reversibly:
i. at constant pressure to a final volume of 0.2 m^3
ii. according to a linear law to a final volume of 0.2 m^3 and a final pressure of 2 bar
iii. according to a law pV = constant to a final volume of 0.1 m^3
iv. according to a law pV^3 = constant to a final volume 0f 0.06 m^3
v. according to a law p=(a/v^2) - (b/v) to a final volume of 0.1 m^3 and a final pressure of 1 bar
where A and B are constants.
Sketch all processes on a p-V diagram.

I have the answer but I need the solution. Thanks

Answer 1

Answer on Question #11567-Physics-Molecular Physics-Thermodynamics

A certain fluid at 10 bar is contained in a cylinder behind a piston, the initial volume being $0.05\mathrm{m}^{\wedge}3$ . Calculate the work done by the fluid when it expands reversibly:

i. at constant pressure to a final volume of $0.2\mathrm{m}^{\wedge}3$

ii. according to a linear law to a final volume of $0.2\mathrm{m}^{\wedge}3$ and a final pressure of 2 bar

iii. according to a law $pV =$ constant to a final volume of $0.1 \, \text{m}^{\wedge} 3$

iv. according to a law $pV^{\wedge}3 =$ constant to a final volume of $0.06 \, \text{m}^{\wedge} 3$

v. according to a law $p = (a / v^{\wedge}2) - (b / v)$ to a final volume of $0.1 \, \text{m}^{\wedge}3$ and a final pressure of 1 bar where A and B are constants.

Sketch all processes on a p-V diagram.

Solution

i.

For reversible non-flow process we have

$$W = \text {s h a d e d a r e a} = \int_ {V _ {1}} ^ {V _ {2}} p d V = p \left(V _ {2} - V _ {1}\right) = 1 0 0 0 k P a \cdot (0. 2 - 0. 0 5) m ^ {3} = 1 5 0 k J.$$

ii.

$$W = \text {s h a d e d a r e a} = \left(\frac {p _ {1} + p _ {2}}{2}\right) | V _ {2} - V _ {1} | = \left(\frac {1 0 + 2}{2}\right) 1 0 ^ {5} | 0. 2 - 0. 0 5 | = 9 0 k J.$$

iii.

$$W = \text{shaded area} = \int_{1}^{2} p \, dV = \int_{1}^{2} \frac{p_{1} V_{1}}{V} \, dV.$$ $$W = p_{1} V_{1} \ln \frac{V_{2}}{V_{1}} = 1000 \cdot 0.05 \cdot \ln \left(\frac{0.1}{0.05}\right) = 34.7 \, \text{kJ}.$$

iv.

$$W = \frac{p_{2} V_{2} - p_{1} V_{1}}{(1 - 3)}.$$

Where $p_2 = p_1 \left(\frac{V_1}{V_2}\right)^3 = 10 \left(\frac{0.05}{0.06}\right)^3 = 5.787 \, \text{bar}$.

$$W = \frac{5.787 \cdot 0.06 - 10 \cdot 0.05}{-2} \cdot 10^{5} = 7640 \, \text{J}.$$

v.

$$W = \int_{1}^{2} \left(\left(\frac{a}{V^{2}}\right) - \left(\frac{b}{V}\right)\right) dV = a \left(\frac{1}{V_{1}} - \frac{1}{V_{2}}\right) + b \ln \frac{V_{1}}{V_{2}}.$$ $$10 \, \text{bar} = \left(\frac{a}{(0.05 \, \text{m}^{3})^{2}}\right) - \left(\frac{b}{0.05 \, \text{m}^{3}}\right); \quad 1 \, \text{bar} = \left(\frac{a}{(0.1 \, \text{m}^{3})^{2}}\right) - \left(\frac{b}{0.1 \, \text{m}^{3}}\right).$$ $$a = 0.04 \, \text{bar} \, \text{m}^{6}; \, b = 0.3 \, \text{bar} \, \text{m}^{3}.$$

Thus,

$$W = 0.04 \, \text{bar} \, \text{m}^{6} \left(\frac{1}{0.05 \, \text{m}^{3}} - \frac{1}{0.1 \, \text{m}^{3}}\right) + 0.3 \, \text{bar} \, \text{m}^{3} \ln \frac{0.05 \, \text{m}^{3}}{0.1 \, \text{m}^{3}} = 19.2 \, \text{kJ}.$$

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