Question

An automobile tire having a temperature of 3.4 ◦C (a cold tire on a cold day) is filled to a gauge pressure of 24 lb/in2 . What would be the gauge pressure in the tire when its temperature rises to 26◦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/in2 . Answer in units of lb/in2 .

Answer 1

Answer:

$27.164 lb/in^2$

Explanation:

We are given that

Gauge pressure at 3.4 degree Celsius,P=$24lb/in^2$

We have to find the gauge pressure in tire when the temperature rises to 26 degree Celsius.

Atmospheric pressure=$14.7lb/in^2$

$P_1=P+14.7=24+14.7=38.7lb/in^2$

K=273+Degree Celsius

$T_1=3.4+273=276.4 K$

$T_2=26+273=299K$

$P_2=\frac{P_1T_2}{T_1}$

$P_2=\frac{38.7\times 299}{276.4}$

$P_2=41.864 lb/in^2$

Gauge pressure in tire when the temperature rises to 26 degree Celsius.=$41.864-14.7=27.164lb/in^2$

Answer 2

Answer:

27.16 lb/in²

Explanation:

initial temperature, T1 = 3.4 °C = 276.4 K

initial gauge pressure, P1 = 24 lb/in²

atmospheric pressure, Po = 14.7 lb/in²

initial absolute pressure, P1' = Po + P1 = 14.7 + 24 = 38.7 lb/in²

final temperature, T2 = 26 °C = 299 K

Let the final gauge pressure is P2.

use the ideal gas equation and the volume is constant.

$\frac{P_{1}'}{T_{1}}=\frac{P_{2}'}{T_{2}}$

$\frac{38.7}{276.4}=\frac{P_{2}'}{299}$

P2' = 41.86 lb/in²

Now the gauge pressure, P2 = P2' - Po = 41.86 - 14.7 = 27.16 lb/in²

Thus, the new gauge pressure is 27.16 lb/in².