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

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 .

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Hồng Cúc 5 years 2021-07-31T05:36:57+00:00 2 Answers 37 views 0

Answers ( )

  1. kimchi2
    0
    2021-07-31T05:38:25+00:00
    Reply

    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.

  2. Adela
    0
    2021-07-31T05:38:55+00:00
    Reply

    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

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