Question

5. A steel-belted radial automobile tire is inflated to a gauge pressure of 1.75 10 Pa when the temperature is 61 °F. Later in the day, the temperature rises to 101 °F. Assuming the volume of the tire remains constant, what is the gauge pressure at the elevated temperature? (Hint: Remember that the ideal gas law uses absolute pressure.] Pa

Answer 1

we know,
Tc = (5/9)*(Tf - 32)

T1 = (5/9)*(61 - 32)

= 16.1 C

= 16.1 + 273

= 289.1 K

T2 = (5/9)*(101 - 32)

= 38.3 C

= 38.3 + 273

= 311.3 K

P1 = P_gauge + P_atm

= 1.75*10^5 + 1.013*10^5

= 2.763*10^5 Pa

P2_gauge = ?

At constant volume, P/T = constant

P2/T2 = P1/T1

P2 = P1*(T2/T1)

= 2.763*10^5*(311.3/289.1)

= 2.975*10^5 Pa

P2_gauge = P2 - P_atm

= 2.975*10^5 - 1.013*10^5

= 1.96*10^5 pa <<<<<<<<<<<---------------------Answer