Question

A heating coil is immersed in a 1 g sample of ${{\text{H}}_2}{\text{O(l)}}$ at 1 atm and ${100^ \circ }{\text{C}}$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 conditions are $1000{\text{Kg}}/{{\text{m}}^3}$ and $0.60{\text{Kg}}/{{\text{m}}^3}$respectively. Magnitude of the work done for the process is:
A.$95.5{\text{ J}}$
B.$100{\text{ J}}$
C.$101.3{\text{ J}}$
D.None of these

Answer 1

To answer this question we must have the knowledge of the formula for work done by a process. From the given densities we can calculate the volume of water in each of the phase and hence put the change in volume in the formula to calculate the work done.
Formula used: ${\text{density}} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}$ and ${\text{W}} = - {\text{P}}\Delta {\text{V}}$ where W is work done and P is pressure and $\Delta {\text{V}}$ is change in volume.

Complete step by step solution:
It is given that 1 g of water was taken initially and the density of water is $1000{\text{Kg}}/{{\text{m}}^3}$. Since $1{\text{Kg}} = 1000g$ and $1{{\text{m}}^3} = 1000{\text{L}}$, we can write density of water as $1000{\text{g}}/{\text{L}}$.
Only $60\%$ of the water is being converted into the gaseous phase and initial amount is 1 g. $60\%$ of 1 g will be:
$1 \times \dfrac{{60}}{{100}} = 0.6{\text{g}}$.
Density of water is already given, that is $1000{\text{g}}/{\text{L}}$. So using the formula we can calculate the volume of liquid that is converted into gaseous phase as
${\text{Volume of liquid converted into gaseous phase}} = \dfrac{{{\text{0}}{\text{.6 g}}}}{{1000{\text{ g}}{{\text{L}}^{ - 1}}}} = 0.0006{\text{ L}}$
Since $0.6{\text{g}}$ of liquid is converted so the same amount of stream must have been produced, that is $0.6{\text{g}}$ of stream is there. Density of stream is $0.60{\text{Kg}}/{{\text{m}}^3}$ or $0.60{\text{g}}/{\text{L}}$ as done previously.
${\text{Volume of stream formed}} = \dfrac{{{\text{0}}{\text{.6 g}}}}{{{\text{0}}{\text{.60 g}}{{\text{L}}^{ - 1}}}} = 1{\text{ L}}$
${\text{W}} = - {\text{P}}\Delta {\text{V}}$ Pressure is 1 atm and change in volume can be calculated by subtracting final minus initial that is by subtracting liquid volume form gaseous volume.
${\text{W}} = - 1{\text{ atm (1}} - 0.0006){\text{L}} = - 0.9994{\text{ Latm}}$
$1{\text{ Latm }} = {\text{ 101}}{\text{.325 Joule}}$
${\text{0}}{\text{.9994 Latm }} = {\text{ 101}}{\text{.3 Joule}}$
Hence C is the correct option.

Note: we know that volume is inversely proportional to the density. This statement is well proved by this question. When we move from liquid to liquid the volume increases because the intermolecular force decreases and particles move away with each other and hence the density will decrease. In the question it is given to us the density of liquid is more than density of gas and in the answer we have calculated that volume of liquid is less than volume of gas.