#### Question

A1-m3 insulated, rigid tank contains air at 800 kPa, 25$^{\circ} \mathrm{C}$. 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 Assuming that the air remaining inside has undergone reversible adiabatic expansion, calculate the mass withdrawn during the process.

#### Solution

Verified#### Step 1

1 of 2We are given following data for Air in rigid tank in reversible adiabatic expansion:

$T_1=298\text{ K}$

$P_1=800\text{ kPa}$

$P_2=150\text{ kPa}$

$V=1\text{ m}^3$

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

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

$k=1.4$

Calculating final temperature:

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

Mass withdrawn during the process is equal to:

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