Why does differentiating Boyle's Law to find chance in volume lead to positive change in V as well as P?

here's my work:

V1*P1=V2*P2 where V1=1000 m^3 P1=5 N/m^2 P2=10 N/m^2 and dP/dT=.05 N/m^2

if V1(dP/dT) + P1(dV/dT)=V2(dP/dT) + P2(dV/dT)

then dV/dT=(V2-V1)*(dP/dT)/(P1-P2)

so the increase in pressure gave decrease in volume but dV/dT comes out positive...where did I go wrong?

1 Answer

  • 1 decade ago
    Favorite Answer

    You can't differentiate BOTH sides. That equation only tells you something if you let one side vary and hold the other side constant.

    Think of it like

    V(0) P(0) = V(t) P(t).

    The left side is a constant value, while the right side involves functions of t. Now this equation is actually telling you something, and since it's true for every t, you could differentiate it.

    The derivative of the left side is zero, since it's a constant.

    0 = V(t) dP/dt + P(t) dV/dt.

    To find dV/dt (I assume when t corresponds to situation P2 and V2), you first need to find V2. You'd do that using the original equation, since you already know V1, P1, and P2.

    Once you find V2, you plug in P2, V2, and dP/dt into the derivative equation to find dV/dt. It ends up being negative, as you'd expect.

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