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• Notes

$\begin{array}{l}{\text { Ten kilograms of } R-134 a \text { till a } 1.595-m 3 \text { weighted piston-cylinder device }} \\ {\text { at a temperature of }-26.4^{\circ} \mathrm{C} \text { . The container is now heated until the temperature is } 100^{\circ} \mathrm{C} \text { . }} \\ {\text { Determine the final volume of the } \mathrm{R}-134 \mathrm{a} \text { . }}\end{array}$

$m=1 k g \quad, \quad V=150 L=0.15 m^{3}$

$P_{1}=2 M P_{a}, T_{2}=40 c^{\circ}, \quad T_1 ? ? \quad P_{2} ? ?$

insenssial:

$\begin{array}{l}{P_{1}=2 M P_{a}} \\ {v_{1}=0.15 \mathrm{m}^{3}} \\ {T_{1}=? ?}\end{array}$$] \rightarrow A-6 \rightarrow T_{1}=395^{\circ}$

final:

$\begin{array}{l}{T_{2}=40 c^{\circ}} \\ {V_{2}=0.15 \mathrm{m}^{3}} \\ {P_{2}=??}\end{array} ] \rightarrow A-4 \longrightarrow P_{2}=7.385 \mathrm{kPa}$

$\text { Ten kilograms of } R-134 \text { a till } a 1.595-m 3$

$\begin{array}{l}{\text { weighted piston-cylinder device at a temperature of }-26.4^{\circ} \mathrm{C} \text { . }} \\ {\text { The container is now heated until the temperature is } 100^{\circ} \mathrm{C} \text { . }} \\ {\text { Determine the final volume of the } \mathrm{R}-134 \mathrm{a} . \mathrm{V}}\end{array}$

$v=\frac{V}{m}=\frac{1.595}{10}=0.1595 \mathrm{m}^{3} / \mathrm{kg}$

$T_{1}=-2 6.4 c^{\circ}$

$P_{1} ?? \longrightarrow A-12 \rightarrow P=100 \mathrm{KPa}$

$\begin{array}{l}{P_{2}=100 \mathrm{kPa}} \\ {T_{2}=100 \mathrm{c}^{\circ}}\end{array}$$] \rightarrow A-13 \rightarrow v_{2}=0.30138 \mathrm{m}^{3} / \mathrm{kg}$

$∴ V_{2}=m v_{2}=10 * 0.30138=3.0138 \mathrm{m}^{3}$

$\begin{array}{l}{\text { Ten kilograms of } R-134 a \text { till a } 1.595-m 3 \text { weighted piston-cylinder device }} \\ {\text { at a temperature of }-26.4^{\circ} \mathrm{C} \text { . The container is now heated until the temperature is } 100^{\circ} \mathrm{C} \text { . }} \\ {\text { Determine the final volume of the } \mathrm{R}-134 \mathrm{a} \text { . }}\end{array}$

$\text {inituel }:-$

$\begin{array}{l}{P_1=1 M P_{A}} \\ {T_{I}=300^{\circ}c}\end{array}$$\} \rightarrow A-6 \longrightarrow {v_{1}}=0.25799 m^3/kg$

final:-

$\begin{array}{l}{P_{2}=1 \mathrm{MP}{\mathrm{a}}} \\ { x_{2}=0.5}\end{array}$$] \longrightarrow{V_{2}}=V_f+X_2V_{fy}$

$=0.001127+0.5(0.19436-0.001127)$

$=0.09775 m^3/kg$

$\Delta V=m\left(V_{2}-V_{1}\right)=0.8(0.09775-0.25799)$

$=-0.1282 \mathrm{m}^{3}$