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Hard half life equation 
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#1
Jul714, 01:43 PM

P: 29

Just recently I decided to buy an ozone generator that can produce 4 grams of ozone / hour. However, I did a little research and found that ozone has a half life of 12.5 hours at 35 Celsius.
For fun, I decided to find an equation that would let me know the amount of ozone left in the room after leaving the machine on for a certain time. I tried for a while but found it to be extremely challenging since new ozone was being added every moment in time and I had to factor this into account. I believe calculus may be needed for this question. However, I was able to find the maximum amount of ozone this machine was able to sustain, which is around 74.1532 by the following equation, so hopefully it might somehow help: 4/(10.5^0.08) Can anyone help me find the equation? 


#2
Jul714, 02:05 PM

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And 74.1532 is supposed to be what units?



#3
Jul714, 09:14 PM

P: 29

The unit is in grams. Any luck on finding the equation?
So far the closest equation i could come up with is: ozone(h)=4(h)4[(10.5^(2/25*h)))+(10.5^(2/25*(h1)))+(10.5^(2/25*(h2))) .... (10.5^(2/25*(hh))] where h=hours and ozone is the amount of ozone is left in the room in grams. The first part, 4(h) is basically the amount of ozone that can be generated if there is no half life. The second part, in brackets [], is the amount that of ozone that is lost due to its half life. However, the accuracy wont be very great since the ozone half life is only calculated once every hour. To make this more accurate I have to somehow find an equation that simplifies the part in the bracket into an equation that calculates the half life in an infinitesimally small amount of time if you get what i mean. Any help would be appreciated 


#4
Jul714, 10:01 PM

P: 29

Hard half life equation
To elaborate, I could use a very small unit of time (i.e nanoseconds) to make it extremely accurate and the equation would become:
ozone(n)=4(3.6*10^12)(n)  4(3.6*10^12) [(10.5^(12(3.6*10^12)*n)))+(10.5^(12(3.6*10^12)*(n1)))+(10.5^(12(3.6*10^12)*(n2))) .... (10.5^(12(3.6*10^12)*(nn))] where n=nanoseconds and ozone is the amount of ozone is left in the room in grams. However, the equation still wouldn't be perfect and I could make it more accurate by using a shorter unit of time. And even if I used the new unit, I could still use a unit that is smaller an so on and so on. I understand that the answer would be more accurate than enough for all practical purpose and I could use excel to calculate this but is there a more elegant way of doing this? 


#5
Jul814, 01:44 AM

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#6
Jul814, 01:58 AM

P: 29

Unfortunately no, the calculus course I took only taught me how to do basic integration and differentiation. I'll probably have to take it in college though.



#7
Jul814, 02:40 AM

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Let ##O## be the amount by weight of ozone in the room at any one time. Naturally, ##O## is a function of time ##t##. The rate of change of ozone is given by ##\frac{dO}{dt}##. Two things contribute to this, production and decomposition. The first occurs at a constant rate of 4g/h. Let's simply denote that ##k##, for greater flexibility. The second happens at a variable rate, just like exponential decay. You can use the same rate formula here as well, i.e. ##\lambda O##, where ##\lambda = \frac{\ln 2}{t_{\frac{1}{2}}}##. Because this is decay and not production, we need to put a negative sign before the rate. If you're unsure what I'm doing here, you should look up radioactive decay in more detail. So now we can form the differential equation as: ##\frac{dO}{dt} = k  \lambda O## We can now use a basic technique called separation of variables to "rearrange" the equation to: ##\frac{dO}{k  \lambda O} = dt## Now you can integrate both parts to arrive at a solution. I'm going to impose bounds that allow a quick solution without needing to worry about constants. ##\int_0^{O(T)}\frac{dO}{k  \lambda O} = \int_0^Tdt## Here, the lower bound is zero because that denotes the initial state when no ozone has been produced and ##O(T)## denotes the ozone concentration at time ##T##. I chose the big T to avoid confusion with the small t that I'm using as the integration variable. Can you perform the integration of both sides? The RHS is trivial, and the LHS should be a familiar form. 


#8
Jul814, 03:19 AM

P: 29

I took a wild stab at it. Not sure if its correct or not.
http://postimg.org/image/ja8hsz93h/ Ok I just checked it after uploading and surely that cant be the answer. There are no variables in the equation at all! Just realized another mistake: e^ln4=4 so the answer would be 0. Can anyone check and see what I did wrong? 


#9
Jul814, 04:01 AM

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A bit of general advice: it's bad practice to substitute your constants in too early. It's much easier to see the form and recognise if it "looks" correct if you leave your constants as symbols till the last step. Could you please clarify if this is homework? I am limited by forum policy in how explicitly I can help you if it is. 


#10
Jul814, 04:11 AM

P: 29

This isn't homework, I'm currently in break and have tons of time to kill. It should also be a good warm up for me to be prepared for college.
Anyways, I tried it again from scratch and the answers looks somewhat sensible. http://postimg.org/image/drm4mtwg1/ 


#11
Jul814, 04:11 AM

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OK, if this isn't homework, then here's how you solve it:
$$\int_0^{O(T)}\frac{dO}{k  \lambda O} = \int_0^Tdt$$ $$\frac{1}{\lambda}[\ln(k\lambda O(T))  \ln k] = T  0$$ $$\ln\frac{k  \lambda O(T)}{k} = \lambda T$$ $$1  \frac{\lambda O(T)}{k} = e^{\lambda T}$$ And hence, $$O(T) = \frac{k}{\lambda}(1  e^{\lambda T})$$ It's instructive to sketch this curve. Note that at ##T=0##, ##O=0##, as we would expect that the initial ozone concentration is zero. But the more interesting thing happens at very large values of ##T##. Let ##T \to \infty##. The exponential term vanishes at the limit. Then: $$\lim_{T \to \infty}O(T) = \frac{k}{\lambda} = \frac{kt_{\frac{1}{2}}}{\ln 2}$$ That's the "limiting" amount of ozone you can produce after a "very long" time. If you plug in the numbers, you'll get a value very close to what you got before, but this is more accurate. 


#12
Jul814, 08:18 AM

P: 428

That's a rather longwinded way to go about it.
We don't need to know ## O ## as a function of time, we just need to note that steady state is reached when the rate of decay equals the rate of production i.e. ## \lambda O_s = k ## so ## O_s = \frac{k}{\lambda} ##. Substituting for ## \lambda ## we have directly $$ O_s = \frac{kt_h}{\ln 2} $$ 


#13
Jul814, 07:38 PM

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#14
Jul914, 05:57 PM

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