Question

A rigid tank of volume 0.5 $m^3$ is initially evacuated . A thin small hole develops in the wall, and air from surrounding at 1 bar, ${21}^{o}C$ leaks in. Final pressure in the tank is 1 bar and final temperature in the tank ${21}^{o}C$. What is the amount of heat transfer between the tank and surroundings?

• A. 50 kJ
• B. 100 kJ
• C. 70 kJ
• D. 150 kJ

Answer 1

The correct option is A 50 kJ


$V=0.5m^3$

$p_1=1bar,p_2=1bar$

$T_1=294k$

Mass of air entering chamber,

$m_i=\frac{PV}{RT}$

$=\frac{{10}^5\times 0.5}{287\times 294}=0.59kg$

Final mass of air in cylinder,

$m_2=m_1$

Now, $dE/dT{]}_{cv}=\dot{Q}+m_i\left[h_1\right]-m_e{h}_e$

On intergrating

$\Rightarrow m_2{u}_2-m_1{u}_1=Q+[m_2-m_1]h_1$

$m_1=0\text{as tank is initially evacuated}$

$m_e=0\text{as no air exit cylinder}$

$\Rightarrow m_2{c}_v{T}_2=Q+m_2{c}_p{T}_1$

$\Rightarrow m_2[c_v-c_p]\times T_2=Q$

$\Rightarrow Q=0.59\times [-0.287]\times 294$

$=-50kJ\text{(-v sign indicates that heat is lost to surroundings)}$.