A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diameter of 1.50 c

Question

A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diameter of 1.50 cm and will “”pop off”” the test tube if a net upward force of 10.0 N is applied to it. To what temperature would you have to heat the trapped air in order to “”pop off”” the stopper? Assume the air surrounding the test tube is always at a pressure of 1.00 atm.

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Tryphena 5 years 2021-07-19T12:23:55+00:00 1 Answers 269 views 1

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  1. Gerda
    0
    2021-07-19T12:24:55+00:00
    Reply

    Answer:

    180° C

    Explanation:

    First we start by finding the area of the stopper.

    A = πd²/4, where d = 1.5 cm = 0.015 m

    A = 3.142 * 0.015² * ¼

    A = 1.767*10^-4 m²

    Next we find the force on the stopper

    F = (P – P•)A, where

    F = 10 N

    P = pressure inside the tube,

    P• = 1 atm

    10 = (P – 101325) * 1.767*10^-4

    P – 101325 = 10/1.767*10^-4

    P – 101325 = 56593

    P = 56593 + 101325

    P = 157918 Pascal

    Now, remember, in an ideal gas,

    P1V1/T1 = P2V2/T2, where V is constant, then we have

    P1/T1 = P2/T2, and when we substitute the values, we have

    101325/(273 + 18) = 157918/ T2

    101325/291 = 157918/ T2

    T2 = (157918 * 291)/101325

    T2 = 453 K

    T2 = 453 – 273 = 180° C

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