Rate of change of pressure wrt volume

Rate of change of the depth of water in a conical tank

A conical tank with vertex down is 8 metres in diameter and 12 metres deep. Water flows into the tank at 10 m³ per minute. Find the rate of change of the depth of the water at the instant when the water is 6 metres deep.

Solution

Ratio between height and radius: h/r = 12/4 → h = h/3

Volume of conical = 1/3 x π x r² x h
                                 = 1/3 x π x (h/3)² x h

                                 = 1/27 x π x h³

dV/dh = 1/9 x π x h²

Rate of change of volume dV/dt = dV/dh x dh/dt

                                                    10  = 1/9 x  π x h² x dh/dt

When the deep of water is 6 cm

                                                    10  = 1/9 x  π x 6² x dh/dt

                                                    dh/dt = 90/(36π) = 5/(2 π)

Problem solving about differential equation

The organiser of sale, which lasted for 3 hours and raised a total £1000, attempted to create a model to represent the relationship between s and t, where £ s is the amount which had been raised at time t hours after the start of the sale. In the model s and t were taken to be continous variables. The organiser assumed that rate of raising money varied directly as the time remaining and inversely as the amount already raised.  

a.     Find the equation of s in terms of t

b.     Find the amount raised during the first hour of sale

c.      Find the rate of raising money one hour after the start of the sale

Solution

First we must make a model for this problem.

The organiser of sale assumed that the rate of raising money raised varied directly as the time remaining which is (3 – t)  and inversely as the amount already raised (s)

Where k is a constant

We arrange the equation

A Level trigonometry question

Larutan penyangga HCOOH dan NaHCOO

        Sebanyak 690 mg HCOOH dan 566 mg NaHCOO dilarutkan dalam air sehingga 100 ml.
        K_a HCOOH = 1,8 x 10^-4.

                a.       Tentukan pH larutan tersebut

                b.      Jika kedalam larutan ditambahkan 1 ml Ba(OH)₂ 0,25 M, berapakah pH larutan sekarang?

.              c.       Jika yang ditambahkan adalah 1 ml H₂SO₄ 0,25 M, berapakah pHnya

Penyelesaian

        a.      Mol HCOOH =0,69/46 =1,5 x 10^-2

              Mol NaHCOO = 0.566/68 = 8.32 x 10^-3

              Karena larutan ini adalah larutan asam lemah dengan basa konjugasinya maka

             [H⁺] = 1,8 x 10^-4 x (1,5 x 10^-2/8,32 x 10^-3) = 3,25 x 10^-4

             pH = - log [H⁺] = - log (3,25 x 10^-4) = 4 – log 3,25 = 3,49

       b.      Ba(OH)₂ yang ditambahkan akan bereaksi dengan komponen asam yaitu HCOOH

Mol Ba(OH)₂ yang ditambahkan = 1 ml x 0,25 M = 0,25 mmol

Mol OH^- = 2 x 0,25 = 0,5 mmol = 5 x 10^-4 mol

Susunan campuran setelah penambahan Ba(OH)₂ dapat dirinci sebagai berikut:

Reaksi	HCOOH (aq) + OH-(aq)       → HCOO-(aq) +  H2O(l)
Awal	1,5 x 10-2      5 x 10-4       8,32 x 10-3
Bereaksi	-5 x 10-4      -5 x 10-4       +5 x10-4               +5 x 10-4
Sisa	1,45 x 10-3                        8,82 x 10-3            +5 x 10-3

Campuran tetap bersifat penyangga asam, karena mengandung HCOOH dan HCOO^-

[H⁺] = 1,8 x 10^-4 x (1,45 x 10^-2/8,82 x 10^-3) = 2,96 x 10^-4

pH = - log (2,96 x 10^-4) = 4 – log 2,96 = 3,53

           c.       H­₂SO₄ yang ditambahkan akan bereaksi dengan komponen basa yaitu HCOO^-

     Mol H₂SO₄ yang ditambahkan = 1 ml x 0,25 M = 0,25 mmol

     Mol H⁺ = 2 x 0,25 = 0,5 mmol = 5 x 10^-4 mol

     Susunan campuran setelah penambahan H₂SO₄ dapat dirinci sebagai berikut:

Reaksi	H+(aq)        + HCOO-(aq)     → HCOOH(aq)
Awal	5 x 10-4       8,32 x 10-3        1,5 x 10-2
Bereaksi	-5 x 10-4      -5 x 10-4           +5 x 10-4
Sisa	7,82 x 10-3       1,55 x 10-2

Campuran tetap bersifat penyangga asam, karena mengandung HCOOH dan HCOO^-

[H⁺] = 1,8 x 10^-4 x (1,55 x 10^-2/7,82 x 10^-3) = 3,57 x 10^-4

     pH = - log (3,57 x 10^-4) = 4 – log 3,57 = 3,48

Dari perhitungan terlihat bahwa larutan penyangga tidak mengalami perubahan secara drastis ketika ditambahkan sedikit asam maupun basa.