Question

The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire when the air temperature in the tire rises to 44°C. Also, determine the amount of air that must be bled off to restore pressure to its original value at this temperature. Assume the atmospheric pressure to be 100 kPa.

Answer 1

Answer:

ΔP=19.76 KPa

$\Delta m=10^{-3}\ gm$

Explanation:

Given that

$T_1= 25C, P_1= 210 \KPa \gauge$

atmospheric pressure = 100 kPa.

So absolute pressure = Atmospheric pressure + gauge pressure

$P_1=210+100\ KPa$ (absolute)

$P_1=310\ KPa$ (absolute)

Here volume of air is constant .We know that for constant volume pressure

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

here

$T_2= 44C$

$T_1=273+25=298K$

$T_2=273+44=317K$

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

$\dfrac{310}{298}=\dfrac{P_2}{317}$

$P_2=329.76\ KPa$ (absolute)

So rise in pressure

$\Delta P=P_1-P_2$

ΔP=329.76-310 KPa

ΔP=19.76 KPa

$m_1=\dfrac{P_1V}{RT_1}$

$m_1=\dfrac{310\times 0.025}{0.287\times 298}$

$m_1=0.090615\ kg$

$m_2=\dfrac{P_2V}{RT_2}$

$m_2=\dfrac{329.76\times 0.025}{0.287\times 317}$

$m_2=0.090614\ kg$

Δm=0.090615 - 0.090614 kg

$\Delta m=10^{-3}\ gm$