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\begin{array}{l}{\text { A gas is contained in a vertical, frictionless piston-cylinder device. The piston has a mass of } 3.2 \mathrm{kg}} \\ {\text { and across-sectional area of } 35 \mathrm{cm} 2 \text { . }} \\ {\text { A compressed spring above the piston exerts a force of } 150 \mathrm{N} \text { on the piston. }} \\ {\text { If the atmospheric pressure is } 95 \mathrm{kPa} \text { , determine the pressure inside the cylinder }}\end{array}

 

$$
P(A)-P_{atm }(A)
$$$$
-F_{sprng}- w= 0
$$

$$
P(A)=P_{atm }(A)
$$$$
+F_{sprng}+ w
$$

\(÷(A) \)

$$
P=P_{a t m}+
$$$$
\left(\frac{F_{sprng}}{A}+\frac{w}{A}\right)
$$

$$
P=P_{a t m}+\frac{F_{sprn}+w}{A}
$$

$$
=9 5+
$$$$
\frac{150*3.2(9.81)}{35 * 10^{-4}}
$$$$
\left(\frac{1}{1000}\right)={147 \mathrm{kPa}}
$$

$$
\begin{array}{l}{\text { Consider the system shown in Fig. Pl- } 86 . \text { If a change of } 0.7 \mathrm{kPa} \text { in the pressure }} \\ {\text { of air causes the brine-mercury interface in the right column to drop }} \\ {\text { by } 5 \mathrm{mm} \text { in the brine level in the right column while the pressure in }} \\ {\text { the brine pipe remains constant, determine the ratio of } \mathrm{A} 2 ! \mathrm{AX}}\end{array}
$$

$$
\text { Before } g-
$$                $$
P_{A_{1}}+\rho_{w} g h{w}+
$$$$
\rho_{\mathrm{Hy}}gh_{hg1}-\rho_{br}gh_{b.1}=P_B
$$

$$
\text { After g- }
$$                $$
P_{A{2}}+\rho_{w} g h{w}
$$$$
+\rho_{H y} g h_{H y 2}
$$$$
-\rho_{br} g h_{b r_{2}}=P_{B}
$$

(After - Before)

$$
P_{A 2}-P_{A 1}+\rho_{H y} g \Delta h_{Hy}
$$$$
- \rho_{b r} g \Delta  h_{br}=0
$$

$$
\frac{P{A1}-P_{ A2}}{\rho_{ w g}}=
$$$$
S G_{H g} \Delta h_{H g}-S G_{b r} \Delta h_{b r}=0
$$

$$
P_{A 2}-P_{A 1}
$$$$
=-0.7 \mathrm{kPs}=-700 \mathrm{N} / \mathrm{m}^{2}=-700{\mathrm{kg}}/ \mathrm{ms}^{2}
$$

$$
\Delta  h_{br }=0.005 \mathrm{m} \longrightarrow
$$$$
\Delta h_{H g}=\Delta h_{H g \ right}+\Delta h_{H \ le f}
$$

$$
=\Delta h_{b r}+\Delta h_{b r} (A2/A1)
$$

$$
=\Delta h_{b r}\left(1+\frac{A{2}}{A{1}}\right)
$$

$$
\frac{700}{1000 * 9.81}=(13.96*(0.005)(1+\frac{A1}{A2})-(1.1)(0.005))
$$

$$
{A{2} / A{1}}=0.134
$$

 

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