Atank contains 80 gallons of pure water. a brine solution with 2 ; b/gal of salt enters at 2 gal/min and the well stirred mixture leaves at the same rate. find the amount of salt in the tank at any time and the time at which the brine leaving will contain 1 lb/ gal of salt? ​

Given:
vi=80 gal
cin=2 lb/gal
qin=2 gal/min
qout=2 gal/min
(since qin=qout, the volume is constant)
assume xo(original amount)=0
cout=x/v=x/80

dx/dt=cinqin-coutqout

dx/dt=(2)(/80)(2)

dx/dt=2[(160-x)/80]

dx/dt=(160-x)/40 (separable de)

dx/(160-x)=dt/40 (integrate)

-ln(160-x)=(1/40)t+c (general solution)

initial condition: xo=0,t=0
-ln(160-0)=(1/40)(0)+c
c=-ln(160)=-5.0752

-ln(160-x)=(1/40)t-5.0752 (particular solution)

a.) x at any time t x(t)

-ln(160-x)=(1/40)t-5.0752

ln[(160-x)^-1]=(0.025)t-5.0752

e^{ln[(160-x)^-1]}=e^(0.025t-5.0752)

(160-x)^-1=e^(0.025t-5.0752)

1/(160-x)=e^(0.025t-5.0752) (isolate x)

1/[e^(0.025t-5.0752)]=160-x

x(t)=160-{1/[e^(0.025t-5.0752)]}

b. if c=1 lb/gal, t=?

from c=x/v

1=x/80
x=80

-ln(160-x)=0.025t-5.0752

-ln(160-80)=0.025t-5.0752
-ln(80)=0.025t-5.0752

0.025t=5.0752-ln(80)

0.025t=0.6932

t=27.728 mins