Question

1.         The closed end of a Boyle’s law crooks tube is shown to the right. The internal diameter of the tube is 4.0 mm. The tube is cylindrical topped by a hemisphere with the same radius as the tube. It is 3.8 cm from the mercury to the bottom of the hemisphere.

What is the volume of gas trapped in this tube in cm³?

2.         4.25 L of air was trapped in a balloon at sea level where the pressure is 1.00 atm. You throw it in your car and drive to red Lodge Montana (at 5500 ft) and then up the Beartooth Pass at 10,006 ft, (See the next page) where atmospheric pressure is 0.703 atm.

            What is the volume of the balloon on top of the pass?

3.         In actuality it is cold on top of this pass. The winter snow typically keeps the pass closed until about Memorial Day and sometimes July 4^th.

            Suppose the balloon was filled on a hot day when T = 28^oC. On top of the pass on a June day is it a cool 3^oC. (I hope you remembered you coat!)

            Considering both the pressure and temperature, what is he balloon’s volume?

Answer 1

Ans. #1. Radius, r = diameter/2 = 0.4 cm/2 = 0.2 cm     ; [1 cm = 10 mm]

Length of cylindrical portion = Total length of tube – length of hemispheres

                                                = 3.8 cm – 2 x (r)

                                                = 3.8 cm – (2 x 0.2 cm)

                                                = 3.4 cm

Now,

Total Volume of Boyle’s law crooks tube =

Volume of cylinder + 2 x volume of hemispheres

= (pi r² h) + 2 x (2/3 pi r³)

= [3.14159 x (0.2 cm)² x 3.4 cm] + [ 2 x (2/3) x 3.14159 x (0.2cm)³)

= 0.42726 cm³ + 0.03351 cm³

= 0.46077 cm³

= 4.6077 x 10^-4 L                             ; [1 L = 10³ cm³]

Assuming the tube be completely filled with air, the volume of air in it = 4.6077 x 10^-4 L

#2. Change in volume at constant temperature is given by Boyle’s Law: For a gas at constant temperature confined in a closed vessel, the product of pressure and volume is a constant. That is, PV = Constant - Temperature kept constant

Applying Boyle’s law for the given case-

P₁V₁ (at sea level) = P₂V₂ (at Beartooth Pass)

Putting the values in above equation-

1.0 atm x 4.25 L = 0.703 atm x V2

or, V2 = (1.0 atm x 4.25 L) / 0.703 atm = 6.05 L

Therefore, volume of balloon on top of pass = 6.05 L , at 28.0⁰C, 0.703 atm

#3.Using Charles’ Law: Volume (V) of a gas in a closed vessel is directly proportional to absolute temperature (T).

That is, V₁ / T₁(sea level) = V₂ / T₂(Beartooth pass)                  - Pressure kept constant

We have, V1 = 6.05 L, at T1 = 28.0⁰C        - from #2

                 T1 = 28.0⁰C          - At sea level

                 T2 = 3.0⁰C

Putting the values in above equation-

            V₂ = (6.05 L / 28.0⁰C) x 3.0⁰C = 0.648 L

Therefore, volume of balloon on top of pass = 0.65 L , at 3.0⁰C, 0.703 atm