A heating coil is immersed in a 100g sample of H2Ol at 1 atm and 100 - C in a closed vessel- In this heating process- 60

The correct option is C 9994 J Given: Mass of the H_2O = 100 g , #160; #160; 1 atm 100^∘C #160; #160; #160; #160; ...

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Standard XII

Chemistry

First Law of Thermodynamics

A heating coi...

Question

A heating coil is immersed in a 100g sample of $H_{2} O (l)$ at 1 atm and $100 \circ C$ in a closed vessel. In this heating process, 60% of the liquid is converted to the gaseous form at constant pressure of 1 atm. The densities of liquid and gaseous water under these conditions are $1000 k g / m^{3}$ and $0.60 k g / m^{3}$ respectively. Magnitude of the work done for the process is

4997 J

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4970 J

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9994 J

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1060 J

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Solution

The correct option is C 9994 J
$Given:-$ Mass of the $H_{2} O = 100 g$ , $1 a t m 100^{\circ} C$
Density, $ρ_{H_{2} O (l)} = 1000 k g / m^{3}$ And $ρ_{H_{2} O (g)} = 0.60 k g / m^{3}$

$Solution:-$

According to question, 60 % conversion took place.

$∴$ Mass of the $H_{2} O (g)$ formed $= 60 g$

$= 0.06 k g$

Volume of $H_{2} O (g)$ formed $= \frac{m a s s}{d e n s i t y} = \frac{0.6}{0.6}$
$= 0.1 m^{3}$

$= 100 L$

Volume of $H_{2} O (l)$ formed $= \frac{m a s s}{d e n s i t y} = \frac{0.6}{1000}$

$= 6 \times 10^{- 4} m^{3}$

$= 0.06 L$

Now

$W = - P (Δ V)$

$= - P (100 - 0.06)$

$= - 1 (99.94) L - a t m$

$= 99.94 \times 100 J$

$= 9994 J$

$Hence the correct option is C$

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