The air in a tire is initially at 380 kpa, 20 c, when the tire volume is 0.120 m^3. as the tire is warmed by the sun, the pressure becomes 450 kpa and the tire volume increases by 5%. determine the final temperature and the mass of air in the tire.

final temperature is 364.32 K

mass of air is 0.5423 kg

Explanation:

given data

pressure p1 = 380 kPa

volume v1 =  0.120 m³

temperature = 20°C = 20 + 273 = 293 K

pressure p2 = 3450 kPa

volume v2 =  5% increase

to find out

final temperature and the mass

solution

we consider here ideal gas

so equation is

p1v1 / t1 = p2v2/t2

put here all value and find t2

and v2 is = 0.120 ( 1 + 0.05) = 0.126 m³

so

t2 = p2v2/ p1v1 × t1

t2 = (450 × 0.126) /  (380 × 0.120)  ×  293

t2 = 364.32 K

so final temperature is 364.32 K

and

mass = p2v2 / Rt2

here R gas constant is 0.287 kJ/kg.K

so

mass = (450 × 0.126) / (0.287 × 364.32 )

mass = 0.5423

so mass of air is 0.5423 kg

0.11 m/s and 0.28 m/s should be correct