A sample of HCI gas is placed in an electric field of 2-5 - 104 NC -1- The dipole moment of each HCI molecule is 3-4 - 10-30 Cm- Find the maximum torque that can act on a molecule-

Given: Electric field intensity, E = 2.5 times; 104 NC minus;1 Dipole moment of an HCl molecule, P = 3.4 times; 10 minus;30 Cm Torque on the molecule, ...

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Standard XII

Chemistry

Electrolysis and Electrolytes

A sample of H...

Question

A sample of HCI gas is placed in an electric field of 2.5 × 10⁴ NC⁻¹. The dipole moment of each HCI molecule is 3.4 × 10⁻³⁰ Cm. Find the maximum torque that can act on a molecule.

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Solution

Given:
Electric field intensity, E = 2.5 × 10⁴ NC⁻¹
Dipole moment of an HCl molecule, P = 3.4 × 10⁻³⁰ Cm
Torque on the molecule,
$Γ = P E \sin θ$
For $Γ_{m a x}$ , putting sinθ = 1, we get
$Γ_{m a x} = P E = 3.4 \times 10^{- 30} 2.5 \times 10^{4} = 8.5 \times 10^{- 26} Nm$

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A sample of HCI gas is placed in an electric field of 2.5 × 104 NC−1. The dipole moment of each HCI molecule is 3.4 × 10−30 Cm. Find the maximum torque that can act on a molecule.

Given: Electric field intensity, E = 2.5 × 104 NC−1 Dipole moment of an HCl molecule, P = 3.4 × 10−30 Cm Torque on the molecule, Γ=PEsinθ For Γmax, putting sinθ = 1, we get Γmax=PE =3.4×10-302.5×104 =8.5×10-26 Nm

A sample of HCI gas is placed in an electric field of 2.5 × 10⁴ NC⁻¹. The dipole moment of each HCI molecule is 3.4 × 10⁻³⁰ Cm. Find the maximum torque that can act on a molecule.

Given:
Electric field intensity, E = 2.5 × 10⁴ NC⁻¹
Dipole moment of an HCl molecule, P = 3.4 × 10⁻³⁰ Cm
Torque on the molecule,
$Γ = P E \sin θ$
For $Γ_{m a x}$ , putting sinθ = 1, we get
$Γ_{m a x} = P E = 3.4 \times 10^{- 30} 2.5 \times 10^{4} = 8.5 \times 10^{- 26} Nm$