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Homework Help: Substance dissolving problem (D.E.)

  1. Jul 23, 2005 #1
    Hi, I have another problem that I couldn't solve.

    Q: The rate at which a certain substance dissolves in water is proportional at the product of the amount undissolved and the difference c1-c2, where c1 is the concentration in the saturated solution and c2 is the concentration in the actual solution. If saturated, 50gm of water would dissolve 20gm of the substance. If 10gm of the substance is placed in 50gm of water and half of the substance is then dissolved in 90 min, how much will be dissolved in 3 hr?

    First of all I don't understand the meaning of " the difference c1-c2, where c1 is the concentration in the saturated solution and c2 is the concentration in the actual solution." What is the concentration of the actual solution? Wouldn't that be 0% all the time since originally no substance was saturated in the 50gm of water?

    I tried to use Newton's Colling Law to solve this problem (since the previous problem was assuming the law to solve it :blushing: )
    Assuming x = the amount of substance left at time t
    [tex]\frac{dx}{dt}=k(c1 - c2)[/tex]
    [tex]\frac{dx}{dt}=k(x - 0)[/tex]
    Separate and integrate the equation I got
    [tex]x = Ce^k^t[/tex]
    Since x(0) = 0 when t=0, c=0; this is where I got stuck for this equation.

    Then I created a different equation:
    [tex]\frac{dx}{dt}=k(x-50) [/tex]
    Integrating the equation we have
    Applying initial condition x(0) = 60 (10gm + 50 gm), when t=0,
    [tex]60 = 50+Ce^0^k[/tex]
    then C = 10
    Appling the additional condition x(1.5)=55 (I think this is wrong because even after the substance is dissolved, the total weight should be still 60gm, right?)
    [tex]55 = 50 + 10e^1^.^5^k[/tex]
    [tex](\frac{1}{2})^\frac{1}{1.5} =e^k[/tex]
    [tex]x(t) = 50 + 10(\frac{1}{2})^\frac{t}{1.5}[/tex]
    Since the problem is asking the amount of substance after 3 min, substitute t with 3,
    I obtained
    [tex]x=50+10(0.5)^2 =52.5[/tex]
    then I did 60-52.5 = 7.5gm
    But the textbook's answer was 7.14gm and I know even though two numbers are close, the way I solved is WRONG. It was just a coincident that I obtained such a close number. However, this is how far I could get with my knowledge and ability.
    I need help.
  2. jcsd
  3. Jul 23, 2005 #2

    George Jones

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    Gold Member

    I took concentration to be (grams of substance dissolved)/(grams of water). The solution is saturated when 20g substance is dissolved in 50g of water, so c1 = 20/50 and c2 = x/50, where x is the number of grams of substance dissolved.

    Using this, I got the same answer as your textbook.

  4. Jul 23, 2005 #3
    Thanks you for your help. My ability of English limited me to throughly comprehend this problem, I guess.
    Yes, I got the same answer, too. <-- OK, I lied. I thought I understood so I wrote this before actually solving the problem again. Shame on me but I couldn't get the same answer by solving
    [tex]\frac{dx}{20-x} = \frac{1}{50}k dt[/tex]
    [tex] x=20 + Ce^\frac{-1}{50}^k^t[/tex]
    Applying initial condition
    so the equation should be [tex]x=20-10e^\frac{-1}{50}^k^t[/tex]
    Applying another condition x=5 when t=1.5
    I got [tex](\frac{3}{2})^\frac{50}{1.5} = e^-^k[/tex]
    Using this I obtained x,
    x = 20???
    What did I do wrong?
    Last edited: Jul 23, 2005
  5. Jul 23, 2005 #4
    first, write down what you know:
    x(t) = amount (gm) substance dissolved
    A = amount (gm) substance originally added to water = 10 gm
    (A - x(t)) = amount UNdissolved (gm) = (10 - x)
    W = water amount (gm) = 50 gm
    c1 = saturated solution concentration (gm solute/gm water) = 20/50
    c2 = actual solution concentration (gm solute/gm water) = x(t)/50
    k = proportionality constant

    now, just write what the problem states:

    [tex] \frac{dx}{dt} \ = \ k \cdot \mbox{(amount UNdissolved)} \cdot \left ( c1 \, - \, c2 \right ) \ = \ k \cdot \left ( 10 - x \right ) \cdot \left ( \frac{20}{50} \, - \, \frac{x}{50} \right ) [/tex]

    you can now SOLVE this diff eq with techniques given in your textbook.
    Last edited: Jul 23, 2005
  6. Jul 23, 2005 #5
    Dear Geosonel:
    Thank you for helping me again! Well, I guess I didn't quite understanding the problem yet. Your explanation helped me a lot to understand what the problem ment. I will try to solve the equation tomorrow since my brain has been used too much today :blushing:
    Please wish me luck!
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