A tire is inflated with air originally at 10 degrees Celsiusand at normal atmosp

Question

A tire is inflated with air originally at 10 degrees Celsiusand at normal atmospheric pressure. During the inflation process,the air is compressed to 25 % of its original volume, and thetemperature is increased to 40 deg C. After the car is driven athigh speed , the tire's air temperature increases to 75 deg C,and the interior volume increases by 3.0 %. What is the new tirepressure (in atm) ? I've worked this problem out, but I am not getting the rightanswer. The right answer is around 3.77 atm i think. Thanks. A tire is inflated with air originally at 10 degrees Celsiusand at normal atmospheric pressure. During the inflation process,the air is compressed to 25 % of its original volume, and thetemperature is increased to 40 deg C. After the car is driven athigh speed , the tire's air temperature increases to 75 deg C,and the interior volume increases by 3.0 %. What is the new tirepressure (in atm) ? I've worked this problem out, but I am not getting the rightanswer. The right answer is around 3.77 atm i think. Thanks.

Explanation / Answer

There are three steps that this air goes through: .   initially         T = 10C = 283 K                      p = 1 atm                       V  is just "Vo " .    it is pumped into thetire       T = 40 C = 313K                                              p2 = ??                                              V = 0.25 Vo .    it is heated during driving    T= 75 C =   348 K                                             p3 = ??                                             V = 1.03 * 0.25 Vo . So now you can use the ideal gas law to equate step 1 and 3(because its the same air). Note that in each case: .            p V = n R T     so      p V / T = nR          and nR is the same for steps 1 and 3. . So...          pV / T   for step 3   = p V / T for step 1 .           p3 * 1.03 * 0.25 Vo / 348   = 1 atm * Vo /283 .          p3 = 1 atm * 348 / 1.03 * 0.25 *283   =    4.77 atm    is the new absolute pressure of the air inthe tire. . If you are asked to find GAUGE pressure then... .         gauge pressure = absolute pressure - 1atm =     3.77 atm . So now you can use the ideal gas law to equate step 1 and 3(because its the same air). Note that in each case: .            p V = n R T     so      p V / T = nR          and nR is the same for steps 1 and 3. . So...          pV / T   for step 3   = p V / T for step 1 .           p3 * 1.03 * 0.25 Vo / 348   = 1 atm * Vo /283 .          p3 = 1 atm * 348 / 1.03 * 0.25 *283   =    4.77 atm    is the new absolute pressure of the air inthe tire. . If you are asked to find GAUGE pressure then... .         gauge pressure = absolute pressure - 1atm =     3.77 atm

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