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$$
\begin{array}{l}{\text { A } 2 \text { -kg mass of helium is maintained at } 300 \mathrm{kPa} \text { and } 27^{\circ} \mathrm{C} \text { in a rigid container. }} \\ {\text { How large is the container, in } \mathrm{m} 3 ?}\end{array}
$$

$$
\begin{array}{l}{3-76 \text { A spherical balloon with a diameter of } 9 \mathrm{m} \text { is filled }} \\ {\text { with helium at } 27^{\circ} \mathrm{C} \text { and } 200 \mathrm{kPa} \text { . Determine the mole number }} \\ {\text { and the mass of the helium in the balloon }}\end{array}
$$

$$
\text { T. } A-1 \longrightarrow R=2.0769
$$

$$
T=27+273=300 \mathrm{k}
$$

$$
P v=m R T \longrightarrow V=\frac{m R T}{P}=\frac{2 * 2.0769*300}{300}
$$

\(∴ {V}=4.154 \mathrm{m}^{3} \)

$$
\begin{array}{l}{\text { A } 1-m 3 \text { tank containing air at } 25^{\circ} \mathrm{C} \text { and } 500 \mathrm{kPa} \text { is connected through }} \\ {\text { a valve to another tank containing } 5 \mathrm{kg} \text { of air at } 35^{\circ} \mathrm{C} \text { and } 200 \mathrm{kPa} \text { . }}\end{array}
$$

$$
\begin{array}{l}{\text { Now the valve is opened, and the entire system is allowed }} \\ {\text { to reach thermal equilibrium with the surroundings, which are at } 20^{\circ} \mathrm{C} \text { . }} \\ {\text { Determine the volume of the second tank and the final equilibrium pressure of air }}\end{array}
$$

$$
A-1 \longrightarrow R=0.287
$$

$$
P v=m R T \rightarrow V_{B}=\frac{m R T_{1}}{P_{1}}
$$

$$
\longrightarrow \frac{5 * 0.287 * 308}{200}={2.21 m^{3}}
$$

$$
m_{A}=\frac{p v}{R T}=\frac{500*1}{0.287 * 298}={5.846}kg
$$

$$
\longrightarrow V=V_{A}+V_{B}=1+2 \cdot 21=3.21 \mathrm{m}^{3}
$$

$$
\longrightarrow M=m_{A}+m_{B}=5.846+5=10.846 \mathrm{kg}
$$

$$
P_{2}=\frac{m R T}{V}=\frac{10.846 * 0.287 * 293}{3.21}=284.1 kPa
$$

$$
\begin{array}{l}{\text { A rigid tank contains an ideal gas at } 1227^{\circ} \mathrm{C} \text { and } 200 \mathrm{kPa} \text { gage. }} \\ {\text { The gas is cooled until the gage pressure reads } 50 \mathrm{kPa} \text { . }}\end{array}
$$

$$
\text { If the atmospheric pressure is } 100 \mathrm{kPa} \text { , determine the final temperature of the gas }
$$

$$
m_{1}=m_ 2
$$

$$
\frac{P_1 V_ 1}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \rightarrow \frac{P_{1}}{T_1} = \frac{P_{2}}{T_{2}}
$$

$$
\longrightarrow \quad T_{2}=T_{1} \frac{P_{2}}{P_1}
$$

$$
=(1227+273) * \frac{(50+100)}{(200+100)}
$$

$$
=750 \mathrm{k}
$$

$$
=477 c^{\circ}
$$

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