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
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 \;psigpV=nRTpV=RTTp=VR=constantT1p1=V2p2p2=T1p1×T2T1=54=285KT2=80=299Kp2=28535×299=36.72psig
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments