Answer to Question #181235 in Mechanical Engineering for Aymn

Question

A l m rigid tank contains propane molecular weight 44) at 100 kPa, 300 K and connected by a valve to another tank of 0.5 m with propane at 250 kPa. 400 K. The valve is opened and the two tanks come to a uniform state at 325 K What is the final pressure?

Answer 1

$R = 0.1886 \frac{kJ}{kgK}$

$m_A = \frac{P_A V_A}{R T_A}= \frac{100 \times 1}{0.1886 \times 300}= 1.7674 kg$

$m_B = \frac{P_B V_B}{R T_B}= \frac{250 \times 0.5}{0.1886 \times 400}= 1.6564 kg$

$V_2= V_A + V_B = 1.5 m^3$

$m_2= m_A + m_B = 3.4243 kg$

$P_2 = \frac{m_2RT_2}{V_2}= \frac{3.4243 \times 0.1886\times 325 }{ 1.5}= 139.9 kPa$