Question

A1-m3 insulated, rigid tank contains air at 800 kPa, 25C^{\circ} \mathrm{C}. A valve on the tank is opened, and the pressure inside quickly drops to 150 kPa, at which point the valve is closed. Assuming that the air remaining inside has undergone reversible adiabatic expansion, calculate the mass withdrawn during the process.

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We are given following data for Air in rigid tank in reversible adiabatic expansion:

T1=298 KT_1=298\text{ K}

P1=800 kPaP_1=800\text{ kPa}

P2=150 kPaP_2=150\text{ kPa}

V=1 m3V=1\text{ m}^3

From Properties table A.5. we can find Air gas constant and heat capacity ratio:

R=0.287 kJ kg KR=0.287\frac{\text{ kJ}}{\text{ kg K}}

k=1.4k=1.4

Calculating final temperature:

T2=T1(P2P1)k1k=298(150800)1.411.4=185 K\begin{align*} T_2&= T_1\cdot \left(\dfrac{P_2}{P_1} \right)^{\dfrac{k-1}{k}}=298\cdot \left(\dfrac{150}{800} \right)^{\dfrac{1.4-1}{1.4}}\\\\ &=185\text{ K} \end{align*}

Mass withdrawn during the process is equal to:

m=m1m2=P1VRT1P2VRT2=80010.28729815010.287185=6.53 kg\begin{align*} m&=m_1-m_2=\dfrac{P_1\cdot V}{R\cdot T_1}-\dfrac{P_2\cdot V}{R\cdot T_2}\\\\ &=\dfrac{800\cdot 1}{0.287\cdot 298}-\dfrac{150\cdot 1}{0.287\cdot 185}\\\\ &=\boxed{6.53\text{ kg}} \end{align*}

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