Answer to Question #300330 in Molecular Physics | Thermodynamics for Tobias Felix

Question

4. An automobile tire initially contains 0.25 lbm of air at 30 psig, 70 OF and have an automatic valve

that releases air whenever the pressure exceeds 32 psig. After driving, the temperature of air inside

rises to 120 OF, (a) what mass of air escapes to the atmosphere? lbm, (b) What is the tire gage

pressure when the remaining air inside has returned to 70 OF? psig

Answer 1

Solution;

Given;

$m_1=0.25lbm=0.113kg$

$P_1=30psig=206.843kPa$

$T_1=70°F=294.261K$

Use the SI units for ease in calculating;

Assuming;

$v_1=v_2$

We know;

$PV=mRT$

$v_1=\frac{0.113×0.287×294.261}{206.843}$

$v_1=v_2=0.04614m^3$

Calculate the mass at 120°F;

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$P_2=\frac{P_1}{T_1}×T_2=\frac{30}{70}×120=51.42$

$P_2=51.42psig=354.528kPa$

Mass at this pressure;

$m_2=\frac{354.528×0.04614}{0.287×322.039}=0.177kg$

The critical mass,at 32psig;

$m_c=\frac{220.632×0.04614}{296.483×0.287}=0.120kg$

Mas that escapes to the atmosphere;

$m_e=m_2-m_c=0.177-0.120$

$m_e=0.057kg$

Answer;0.126lbm

(b)

$\frac{P_1}{T_1}=\frac{P_3}{T_3}$

Therefore;

$PV=mRT$

$P=\frac{mRT}{v}$

$P=\frac{0.124×0.287×294..261}{0.04614}$

$P=226.965kPa$

$P=32.918psig$