Asked by jessie
A mixture contains two chemicals, A and B, with masses mA and mB. The total mass of the mixture, m, remains constant. B is converted into A at a rate that
is inversely proportional to the mass of A and directly proportional to the square of the mass of B in the mixture at any time t.
Form a differential equation and
find a general expression relating the mass of A to the time t.
ive got dma/dt = mb^2/ma
is this right and how do i do the next bit
is inversely proportional to the mass of A and directly proportional to the square of the mass of B in the mixture at any time t.
Form a differential equation and
find a general expression relating the mass of A to the time t.
ive got dma/dt = mb^2/ma
is this right and how do i do the next bit
Answers
Answered by
FredR
dmA/dt = cmB^2/mA
where c is a constant
mB=m-mA so
dmA/dt = c (m-mA)^2/mA
= (c/mA)(mA^2-2mmA+m^2)
= cmA-2cm+cm^2/mA
integrate with respect to t
where c is a constant
mB=m-mA so
dmA/dt = c (m-mA)^2/mA
= (c/mA)(mA^2-2mmA+m^2)
= cmA-2cm+cm^2/mA
integrate with respect to t
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