An automobile tire is inflated to 32 psig pressure at 50 F. After being driven the temperature rise to 75F. Determine the final gage pressure assuming the volume remains constant.
Process is isochoric
Ideal Gas Law
pV=nRTpV=RTpT=RV=constantp1T1=p2T2p2=p1×T2T1T2=75 F=297 KT1=50 F=283 Kp2=32×297283=33.58 psigpV=nRT \\ pV=RT \\ \frac{p}{T} = \frac{R}{V} = constant \\ \frac{p_1}{T_1} = \frac{p_2}{T_2} \\ p_2 = \frac{p_1 \times T_2}{T_1} \\ T_2 = 75 \; F = 297 \;K \\ T_1 = 50 \;F = 283 \;K \\ p_2 = \frac{32 \times 297}{283} = 33.58 \;psigpV=nRTpV=RTTp=VR=constantT1p1=T2p2p2=T1p1×T2T2=75F=297KT1=50F=283Kp2=28332×297=33.58psig
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