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

Answer 1

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

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

        $P_G = 2.925 *10^{5} \ Pa$