The pressure in an automobile tire depends on the temperature of the air in the

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 310 kPa absolute. 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 50 °C. Also determine the amount of air that must be bled off to restore the pressure to its original value at this temperature.

Explanation / Answer

{Pressure #1, Absolute} = P1 = 210 kPa + 100 kPa = 310000Pa {Pressure #2, Absolute} = P2 = ??? {Volume Of Tire, Constant} = V = 0.025 m^3 {Temperature #1} = T1 = 25degC = (25 + 273.15) degK {Temperature #2} = T2 = 50degC = (50 + 273.15) degK For temperature increase at constant volume, we have: P2 = {T2/T1}*P1 = = {(50 +273.15)/(25 + 273.15)}*(310000) = 335994 Pa Thus, pressure increase caused by temperature increase is obtained from: ----> {P2 - P1} = (335994) -(310000) = = 25994 Pa = 25.994 kPa ******************************************************************************** NEXT PART :::: In order to restore original tire pressure, someair must be removed from the tire. Amount of air requiring removal is determinedon the basis of number of moles of air in the tire. Number of moles "n" of air present in the tire at 50 degC andnew pressure of (335994 Pa) isgiven by the ideal gas law to be: n = P2*V/{R*T2} = = (335994)*(0.025)/{(8.314)*(50 + 273.15)} = 3.1265 moles In order to reduce tire pressure to original value of(310000 Pa) while maintaining new temperature of 50 degC, the number of moles ofair must be reduced to "nnew", where: nnew = P1*V/{R*T2} = = (310000)*(0.025)/{(8.314)*(50 +273.15)} = 2.8846 moles Thus, amount of air needing removal to restore original pressureis: {Air RemovalAmount To Restore Pressure} = n -nnew = = (3.1265moles) - (2.8846 moles) = 0.2419 moles

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