Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is , density of liquid is and is its latent

Question

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is $T$ , density of liquid is $ρ$ and $L$ is its latent heat of vaporization. [2012]

Smart Achievers · Accepted Answer

(4)

When radius is decreased by $ρ$ ,

Decrease in surface energy = Heat lost as latent heat

$(4 π R^{2} Δ R ρ) L = 4 π [R^{2} - {(R - Δ R)}^{2}] T$

$\Rightarrow ρ R^{2} Δ R L = T [R^{2} - R^{2} + 2 R Δ R - Δ R^{2}]$

$\Rightarrow ρ R^{2} Δ R L = T (2 R Δ R) [Δ R is very small]$

$\Rightarrow R = \frac{2 T}{ρ L}$

Solution

Q.
Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible The surface tension is $T$ , density of liquid is $ρ$ and $L$ is its latent heat of vaporization. [2012]

1 $\frac{ρ L}{T}$

2 $\sqrt{\frac{T}{ρ L}}$

3 $\frac{T}{ρ L}$

4 $\frac{2 T}{ρ L}$

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Solution

Q. Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible The surface tension is T, density of liquid is ρ and L is its latent heat of vaporization. [2012]

1 ρLT

2 TρL

3 TρL

4 2TρL