Question #246998

An automobile tire is inflated to 35 psig at 54 oF. After being driven, the temperature rise to 80 oF. Determine the final gage pressure assuming volume remains constant


Expert's answer

Assuming ideal gas low

Process is Isochoric

pV=nRTpV=RTpT=RV=constantp1T1=p2V2p2=p1×T2T1T1=54=285  KT2=80=299  Kp2=35×299285=36.72  psigpV=nRT \\ pV = RT \\ \frac{p}{T} = \frac{R}{V} = constant \\ \frac{p_1}{T_1} = \frac{p_2}{V_2} \\ p_2 = \frac{p_1 \times T_2}{T_1} \\ T_1 = 54 =285 \;K \\ T_2 = 80 =299 \;K \\ p_2= \frac{35 \times 299}{285} = 36.72 \;psig


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