Answer to Question #230596 in Molecular Physics | Thermodynamics for Unknown346307

Question

. An air receiver of volume 5.5 m3 contains air at 16 bar and 42°C. A valve is

opened and some air is allowed to blow out to atmosphere. The pressure of the air in the receiver

drops rapidly to 12 bar when the valve is then closed.

Calculate the mass of air which has left the receiver

Answer 1

Gives

$$p_1=16bar\\p_2=12bar\\T_1=42°c=273+42=315K\\V_1=V_2=5.5$$

$R=0.287\times10^{3}$

$$m_1=\frac{p_1V_1}{RT_1}=\frac{16\times10^{5}\times5.5}{0.287\times10^3\times315}=97.34kg$$

We know that

Undergoes reversible Adiabatic process

$(\frac{P_2}{P_1})^{\frac{\gamma-1}{\gamma}}=\frac{T_2}{T_1}$

$T_1\times(\frac{P_2}{P_1})^{\frac{\gamma-1}{\gamma}}={T_2}$

Put value

$315\times(\frac{12}{16})^{\frac{1.4-1}{1.4}}={T_2}$

$T_2=315\times(\frac{12}{16})^{(\frac{2}{7})}$

$T_2=315\times{(\frac{12}{16})}^{0.287}=290K$

$$m_2=\frac{p_2V_2}{RT_2}=\frac{12\times10^{5}\times5.5}{0.287\times10^3\times290}=79.29kg$$

Mass which left receiver

$$m=m_1-m_2=97.34-79.29=18.05kg$$