I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!
If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:
Hello,Originally Posted by wombat1
the process could be best described by an exponential funktion:
where a is the value at the beginning
b (the base) is the constant of proportionality you are looking for
t is the time.
Put in the values you know:
Solve for b:and you'll get: b = 0.5.
Greetings
EB
Here is one way.
"Decomposes at a rate proportional to amount of substance at present"
Let present be when time is zero.
So,
amount of substance at present = "P at t=0" = Po
amount of substance at time t = "P at t=t" = P of t = P(t)
P(t) = (Po)*e^(kt) -------------exponential function
Given:
at time t=2hrs,
P(t) = P(2) = 10gm
Po = 40gm at anytime.
So,
10 = 40*e^(k*2)
10/40 = e^(2k)
1/4 = e^(2k)
ln(1/4) = ln[e^(2k)] -----(i)
ln(0.25) = (2k)*ln(e)
ln(0.25) = 2k
k = (1/2)*ln(0.25)
k = -0.693147181 -------------------answer.
Or, from (i),
ln(1/4) = ln[e^(2k)] -------(i)
ln(1) -ln(4) = (2k)*ln(e)
0 -ln(2^2) = (2k)*1
-2ln(2) = 2k
k = -ln(2) ---------------(ii)
k = -(0.693147181)
k = -0.693147181
Same as above, hence, k = -ln(2).
Then,
k = -0.693147181 = -ln(2) ----------if you are using the "e".
If you are not using the "e",
P(t) = (Po)*e^(kt)
P(t) = (Po)*[e^(k*t)]
P(t) = (Po)*[(e^K)^t]
To not to use "e", let (e^k) = a = constant of proportionality.
So,
P(t) = (Po)*[a^t]
Plug in the known values,
10 = 40[a^2]
10/40 = a^2
1/4 = a^2
a = 1/2
a = 0.5 ---------------answer.
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