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]

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

b. lifting dumbbell weight at 20-degree angle from the body after a 45-degree angle