An automobile tire having a temperature of
−2.6°C (a cold tire on a cold day) is filled to a
gauge pressure of 24 lb/in^2
What would be the gauge pressure in the
tire when its temperature rises to 43◦C? For
simplicity, assume that the volume of the tire
remains constant, that the air does not leak
out and that the atmospheric pressure remains constant at 14.7 lb/in^2
Answer in units of lb/in^2

Psm = 30.66 [Psig]

Explanation:

To solve this problem we will use the ideal gas equation, recall that the ideal gas state equation is always worked with absolute values.

P * v = R * T

where:

P = pressure [Pa]

v = specific volume [m^3/kg]

R = gas constant for air = 0.287 [kJ/kg*K]

T = temperature [K]

For the initial state

P1 = 24 [Psi] + 14.7 = 165.47[kPa] + 101.325 = 266.8 [kPa] (absolute pressure)

T1 = -2.6 [°C] = - 2.6 + 273 = 270.4 [K] (absolute Temperature)

Therefore we can calculate the specific volume:

v1 = R*T1 / P1

v1 = (0.287 * 270.4) / 266.8

v1 = 0.29 [m^3/kg]

As there are no leaks, the mass and volume are conserved, so the volume in the initial state is equal to the volume in the final state.

V2 = 0.29 [m^3/kg], with this volume and the new temperature, we can calculate the new pressure.

T2 = 43 + 273 = 316 [K]

P2 = R*T2 / V2

P2 = (0.287 * 316) / 0.29

P2 = 312.73 [kPa]

Now calculating the manometric pressure

Psm = 312.73 -101.325 = 211.4 [kPa]

And converting this value to Psig

Psm = 30.66 [Psig]

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.