A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340 kPa. The bot

Question

A two-liter bottle is one-fourth full of water and three-quarters full of air. The air in the bottle has a gage pressure of 340 kPa. The bottle is turned upsidedown and the cap is released so that the water is rapidly forced out of the bottle. If the air in the bottle undergoes an adiabatic pressure change, what is the pressure in the bottle when the bottle is five-sixths full of air

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Thạch Thảo 3 years 2021-07-25T19:41:38+00:00 1 Answers 11 views 0

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  1. lethu
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    2021-07-25T19:43:29+00:00
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    Answer:

    The value is     P_G = 2.925 *10^{5} \ Pa

    Explanation:

    From the question we are told that

       The  volume of the bottle is  v = 2 \ L = 2 * 10^{-3} \ m^3

       The gauge pressure of the air is P_g = 340 \ kPa = 340 *10 ^{3} \ Pa

       

    Generally the volume of air before the bottle is turned upside down is  

          V_a = \frac{3}{4} * V

           V_a = \frac{3}{4} * 2 *10^{-3}

           V_a = 0.0015 \ m^3 }

    Generally the volume air when the bottle is turned upside-down is

          V_u = \frac{5}{6} * 2 *10^{-3}

           V_u = 0.00167 \ m^3

    From the the mathematical relation of adiabatic process we have that

         P_g * V_a^r = P_G * V_u^r

    Here r is a constant with  a value  r =  1.4

    So

          340 *10 ^{3} * 0.0015^{1.4} = P_G * 0.00167^{1.4}

            P_G = 2.925 *10^{5} \ Pa

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