An initially deflated and flat balloon is connected by a valve to a storage tank containing helium gas at 1 MPa at ambient temperature of 20 degrees C. The valve is open and the balloon is inflated at constant pressure of 100 kPa (atmospheric pressure) until it becomes spherical at D1 = 1m. If the balloon is larger than this, the balloon material is stretched giving a pressure inside as: P = Po + C(1-(D1/D))(D1/D)

The balloon is slowly inflated to a final diameter of 4m, at which point the pressure inside is 400 kPa. The temperature remains constant at 20 degrees C. Determine the work done during the overall process.

14660.75 kJ.

Explanation:

We are given from the question above that the pressure inside is represented by this equation:

P = Po + C(1-(D1/D))(D1/D).

Thus, P = 100 + C (1 - 1/4) (1/4). Therefore, P  which is = 400 is;

400 = 100 + C (1 - 1/4) (1/4).

Hence, C = 1600. And the equation can be written as P = 100 + 1600 (1 - (D1/D))(D1/D).

The work done during the entire process, w = (work done during the process 1 to 2) + (work done duringbthe process 2 to 3 ).

Therefore, work done during the process 1 to 2 = P(V2 + V1) = 100( π/6 - 0).

{NB: V2 is gotten from π/6 × D^3. Where D^3 = 1. }

Hence, work done during the process 1 to 2 = 52.36 kJ.

So, the work done duringbthe process 2 to 3 ) is going to be the integration of PdV. Where P = 100 + 1600 (1 - (D1/D))(D1/D) and dV = π/2 D^2 dD. Taking the upper and the lower limit as 4 and 1 respectively.

After the integration (kindly check attached picture for the process) the work done duringbthe process 2 to 3  = 14.61 × 10^3 kJ.  

Therefore, The work done during the entire process, w = (work done during the process 1 to 2) + (work done duringbthe process 2 to 3 ).

The work done during the entire process, w = 52.36 kJ + 14.61 × 10^3 kJ.   = 14660.75 kJ.

cytokinesis is the splitting of a cell membrane to form 2 identical daughter cells. it's similar to mitosis as both cells have identical copies of dna and cytokinesis is a step in the mitotic cycle.

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l