Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given that its conductivity is 0.80 ohm-1cm-1.
for resistance we have R=ρ l/a
thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ
σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
=0.26 Ω-1
Answer: \(0.266 ohm^{-1}\)
Explanation: Specific conductance or conductivity is the conductance for unit volume of a solution.
\(\kappa=C\times \frac{l}{a}\)
where,
\(\kappa\) =Specific conductance or Conductivity=\(0.80ohm^{-1}cm^{-1}\)
C = Conductance = ?
l = length= 9.0 cm
a = area = 3.0\(cm^2\)
\(\frac{length}{area}\)= cell constant =\(\frac{9.0cm}{3.0cm^2}=3cm^{-1}\)
\(0.80ohm^{-1}cm^{-1}=C\times 3cm^{-1}\)
\(C=0.266 ohm^{-1}\)