A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is

Question

A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?

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Minh Khuê 5 years 2021-07-14T20:08:25+00:00 1 Answers 22 views 0

Answers ( )

  1. thienthanh
    0
    2021-07-14T20:10:08+00:00
    Reply

    Answer:

    7.8% of the original volume.

    Explanation:

    From the given information:

    Temperature T_1 = 22° C = 273 + 22 = 295° C

    Pressure P_1 = 240 kPa

    Temperature T_2 = 45° C

    At initial temperature and pressure:

    Using the ideal gas equation:

    P_1V_1 =nRT_1

    making V_1 (initial volume) the subject:

    V_1 = \dfrac{nRT_1}{P_1}

    V_1 = \dfrac{nR*295}{240}

    Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

    the final volume V_2 can be computed as:

    V_2 = \dfrac{nR*318}{240}

    Now, the change in the volume ΔV =  V₂ – V₁

    \Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}

    \Delta V = \dfrac{23nR}{240}

    ∴

    The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

    = \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}

    = {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}

    = 0.078

    = 7.8% of the original volume.

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