The Student Room Group

C4 Further Differentiation small changes

Please check I am right

Question:
if the pressure and volume of a gas are ρ \rho and ν \nu then Boyle's law states ρν=constant(k) \rho\nu = \mathrm{constant (k)} . if δρ \delta \rho and δν \delta \nu denote corresponding small changes ρ \rho and ν \nu express δρρ \frac{\delta \rho}{\rho} in terms of δνν \frac{\delta \nu}{\nu}


My attempt:
theory being used to answer this question:


δyδx≈dydx \frac{\delta y}{\delta x} \approx \frac{\mathrm{d} y}{\mathrm{d} x}
thus
δy≈dydx×δx \delta y \approx \frac{\mathrm{d} y}{\mathrm{d} x}\times \delta x

thus

δρδν≈dρdν \frac{\delta \rho}{\delta \nu} \approx \frac{\mathrm{d} \rho}{\mathrm{d} \nu}
thus
δρ≈dρdν×δν \delta \rho \approx \frac{\mathrm{d} \rho}{\mathrm{d} \nu}\times \delta \nu


ρν=constant(k) \rho\nu = \mathrm{constant (k)}
ρ=constant(k)ν−1 \rho = \mathrm{constant (k) \nu^{-1}}
dρdν=−constant(k)ν−2 \frac{\mathrm{d} \rho}{\mathrm{d} \nu} = - \mathrm{constant (k)\nu^{-2}}

δρ≈−constant(k)ν−2×δν \delta \rho \approx - \mathrm{constant (k)\nu^{-2}} \times \delta \nu

δρρ≈−constant(k)ν−2constant(k)ν−1×δν\frac{\delta \rho}{\rho} \approx -\frac{\mathrm{constant (k)\nu^{-2}}}{constant (k)\nu^{-1}}\times \delta\nu
δρρ≈−δνν \frac{\delta \rho}{\rho} \approx - \frac{\delta\nu}{\nu}

I would like to know if the method used to find the answer

δρρ=−δνν \frac{\delta \rho}{\rho} = - \frac{\delta\nu}{\nu} is right

Thank you
Reply 1
Avatar for mqb2766
mqb2766
21
Looks ok to me
Reply 2
Avatar for bigmansouf
bigmansouf
OP
20
Original post by mqb2766
Looks ok to me

thank you

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you