Question

The air pressure inside the tube of a car tire is 353 kPa at a temperature of 14.0 °C. What is the pressure of the air, if the temperature of the tire increases to 63.0 °C? Assume that the volume of the tube doesn't change.

What is the air pressure inside the tube, if the volume of the tube is not constant, but it increases from 21.0 l to 21.9 l during the warming process described above?

Answer 1

ans)

from above data that

given that

Initial pressure P = 353 Kpa

Initial temperature T = 14 ^o C

                               = 14 + 273

                               = 287 K

Final temperature T ' = 63^o C

                                = 63+ 273

                                = 336 K

At constant volume , P ' / P = T ' / T

From this P ' = (T' /T ) P

                     = 413.3 kPa

Initial volume V =  21.0 l

Final volume V ' = 21.9 l

From the relation PV = nRT

                     PV /T = constant

                  P'V'/T' = PV/T    

From this P ' = PVT' / V'T

                     = (353*21.0*336) /(21.9 * 287 )

                     = 396.3 kPa