Question

A fluid at 0.7 bar occupying 0.09m3 is compressed reversibly to a pressure of 3.5 bar according to law pvn = constant. The fluid is then heated reversibly at constant volume until the pressure is 4 bar; the spesific volume is then 0.05m3 / kg. A reversible expansion according to a law pv2 = constant restores the fluid to its initial state. Sketch the cycle to a p-v diagram and calculate: i) The mass of fluid present (THIS is the one I cannot seem to figure out.. basics be damned) ii) the value of n in the first process. iii) the net work of the cycle.

Answer 1

Answer on Question #70942, Physics / Molecular Physics | Thermodynamics

Question. A fluid at 0.7 bar occupying $0.09m^3$ is compressed reversibly to a pressure of 3.5 bar according to law $pV^n = \text{constant}$ . The fluid is then heated reversibly at constant volume until the pressure is 4 bar; the specific volume is then $0.05m^3/kg$ . A reversible expansion according to a law $pV^2 = \text{constant}$ restores the fluid to its initial state. Sketch the cycle to a $p - V$ diagram and calculate:

i) The mass of fluid present (THIS is the one I cannot seem to figure out.. basics be damned).

ii) the value of $n$ in the first process.

iii) the net work of the cycle.

Given.

$p_1 = 0.7$ bar $= 70000$ Pa, $V_{1} = 0.09m^{3}$ $pV^n =$ constant;

$p_2 = 3.5$ bar $= 350000$ Pa;

$p_3 = 4$ bar $= 400000$ Pa;

$V_{2} = V_{3}$

$pV^2 =$ constant;

$v = 0.05m^3 /kg.$

Find.

$m,n,A - ?$

Solution.

i) For $1 - 2$

$$p _ {1} V _ {1} ^ {n} = p _ {2} V _ {2} ^ {n}.$$

Since $V_{2} = V_{3}$ then for 3 - 1 we have

$$p _ {3} V _ {3} ^ {2} = p _ {3} V _ {2} ^ {2} = p _ {1} V _ {1} ^ {2} \rightarrow V _ {2} = V _ {3} = V _ {1} \sqrt {\frac {p _ {1}}{p _ {3}}} = 0. 0 9 \sqrt {\frac {0 . 7}{4}} = 0. 0 3 7 6 m ^ {3}.$$

So

$$m = \frac {V _ {3}}{v} = \frac {0 . 0 3 7 6}{0 . 0 5} = 0. 7 5 3 k g.$$

ii) From (1)

$$\frac {p _ {1}}{p _ {2}} = \left(\frac {V _ {2}}{V _ {1}}\right) ^ {n} \rightarrow \ln \left(\frac {p _ {1}}{p _ {2}}\right) = \ln \left(\frac {V _ {2}}{V _ {1}}\right) ^ {n} \rightarrow \ln \left(\frac {p _ {1}}{p _ {2}}\right) = n \ln \frac {V _ {2}}{V _ {1}} \rightarrow n = \frac {\ln \left(\frac {p _ {1}}{p _ {2}}\right)}{\ln \frac {V _ {2}}{V _ {1}}} = \frac {\ln \frac {0 . 7}{3 . 5}}{\ln \frac {0 . 0 3 7 6}{0 . 0 9}} = 1. 8 4.$$

iii) The net work of the cycle

$$A = A _ {1 - 2} + A _ {2 - 3} + A _ {3 - 1}$$ $$\begin{array}{l} A _ {1 - 2} = \int_ {V _ {1}} ^ {V _ {2}} p d V = \int_ {V _ {1}} ^ {V _ {2}} \frac {p _ {1} V _ {1} ^ {1 . 8 4}}{V ^ {1 . 8 4}} d V = p _ {1} V _ {1} ^ {1. 8 4} \int_ {0. 0 9} ^ {0. 0 3 7 6} \frac {1}{V ^ {1 . 8 4}} d V = 7 0 0 0 0 \cdot 0. 0 9 ^ {1. 8 4} \cdot \frac {V ^ {- 1 . 8 4 + 1}}{- 1. 8 4 + 1} \left| \begin{array}{l} 0. 0 3 7 6 \\ 0. 0 9 \end{array} \right. = \\ = \frac {8 3 3 . 5}{- 0 . 8 4} \cdot \left(\frac {1}{V ^ {0 . 8 4}} \Bigg | \begin{array}{c} 0. 0 3 7 6 \\ 0. 0 9 \end{array} \right) = \frac {8 3 3 . 5}{- 0 . 8 4} \cdot \left(\frac {1}{0 . 0 3 7 6 ^ {0 . 8 4}} - \frac {1}{0 . 0 9 ^ {0 . 8 4}}\right) \approx - 8 1 0 8 J. \\ A _ {2 - 3} = \int_ {V _ {2}} ^ {V _ {3}} p d V = 0. \\ \end{array}$$ $$\begin{array}{l} A _ {3 - 1} = \int_ {V _ {3}} ^ {V _ {1}} p d V = \int_ {V _ {3}} ^ {V _ {1}} \frac {p _ {1} V _ {1} ^ {2}}{V ^ {2}} d V = p _ {1} V _ {1} ^ {2} \int_ {0. 0 3 7 6} ^ {0. 0 9} \frac {1}{V ^ {2}} d V = 7 0 0 0 0 \cdot 0. 0 9 ^ {2} \cdot \frac {V ^ {- 2 + 1}}{- 2 + 1} \left| \begin{array}{l} 0. 0 9 \\ 0. 0 3 7 6 \end{array} \right. = \\ = \frac {5 6 7}{- 1} \cdot \left(\frac {1}{V} \Bigg | \begin{array}{c} 0. 0 9 \\ 0. 0 3 7 6 \end{array} \right) = \frac {5 6 7}{- 1} \cdot \left(\frac {1}{0 . 0 9} - \frac {1}{0 . 0 3 7 6}\right) \approx 8 7 8 0 J. \\ \end{array}$$

Finally

$$A = - 8 1 0 8 + 0 + 8 7 8 0 = 6 7 2 J.$$

Answer. $m = 0.753\, kg; n = 1.84; A = 672\, J.$

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