Before beginning a long trip on a hot day, a driver inflates an automobile tire

Question

Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of 1.8 atm at 282K. at the end of the trip the gauge pressure in the tire has increased to 2.4atm. assuming the volume of the air inside the tire has remained constant, what is its temperature at the end of the trip?answer in units of K. THis answer is : 342.4285714K gotten by:  a) Assuming that the volume has remained constant, what is thetemperature of the air inside the tire? {New Temp} = {OldTemp}*{New Absolute Pressure}/{Old AbsolutePressure} = = (282degK)*(1.0 + 2.4 abs atm)/(1.0 + 1.8 absatm) = 342.4285714 K.

NOW I need to know: Air is released from the tire during a short time interval, so that the temperature remains at the value found in part 1. Assume that the amound of air released is small enough for the tires volume to be treated as constant. What is the quantity of air (as a fraction of Ni, the initial number of particles) must be released from the tire so that the pressure returns to its initial value? answer in units Ni.

Explanation / Answer

(1)

now rearrange as p/n = RT/V, since only p and n change.
Initially, p/n = 3.4atm/n?
Finally, p/n = 2.8atm/n?
Since these initial states both = RT/V,
n?/ n?= 2.8/3.4 = 0.82, so 18% of the mass should be released.

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