Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the end of the trip the gaugepressure has increased to 3.20 atm.

(a) Assuming that the volume has remainedconstant, what is the temperature of the air inside the tire?
1 K
(b) What percentage of the original mass of air in the tire shouldbe released so the pressure returns to the original value? Assumethe temperature remains at the value found in (a), and the volumeof the tire remains constant as air is released.

The value is the temperature of the air inside the tire 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet = 3.2 + 1 = 4.2 atm

Temperature at inlet = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

value of  the original mass of air in the tire should be released is  

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

multiply 2kg by latent heat of vap

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.