Question

Work done in vaporization of one mole of water at 373K against the pressure of 1atm is approximately :
A.$\, - 3100J\,$
B.$\, - 31.20J\,$
C.$\, - 20.2J\,$
D.$\, + 3100J\,$

Answer 1

Hint: In thermodynamics, there is no work in a system, and work is an operation performed by or on a system. Gases can work against continuous exterior pressure by expansion or compression. Gas work is often also called pressure-volume or $\,PV\,$ work.
Formula used:
$PV = nRT\,$
Where, $\,P = \,$Pressure
$\,V = \,$Volume
$\,n = \,$number of moles
$\,R = \,$Universal gas constant
$\,T = \,$absolute temperature
$\,W = - {P_{ext}} \times \Delta V$
Where $\,W = \,$Work
$\,{P_{ext}} = \,$External Pressure
$\,\Delta V = \,$Change in volume

Complete step by step solution:
Let us first calculate volume of water before vaporization;
$V = \dfrac{m}{d}\,$
Where, $\,V = \,$volume, $\,m = \,$mass and $\,d = \,$density
Here, we have one mole of water which is equal to $\,18g\,$
Therefore, $\,V = \dfrac{{18g}}{{1g/ml}}\,$
As the density of one mole of water is equal to $\,1g/ml\,$at $\,373K\,$
Therefore by solving the equation we get, $\,{V_1} = 18ml\, = 0.018l\,$
Now, let’s calculate $\,{V_2}\,$ which is the volume of water after vaporization;
$P{V_2} = nRT\,$
Where, $\,P = \,$ Pressure
$\,{V_2} = \,$ Volume after vaporization
$\,n = \,$ number of moles
$\,R = \,$ Universal gas constant
$\,T = \,$ absolute temperature
From this we get;
$\,{V_2} = \dfrac{{nRT}}{P}\,$ as per ideal gas equation mentioned above
$\,n = 1,\,R = 0.0821L.atm/mol.K,\,T = 373K,P = 1atm\,$
Substituting them in the equation we get;
$\, \Rightarrow \dfrac{{1 \times 0.0821 \times 373}}{1} = 30.6l\,$
Here $\,{V_1}\,$ is negligible with respect to $\,{V_2}\,$, So we can ignore $\,{V_1}\,$and therefore, $\,\Delta V = 30.6l\,$
Now, let us calculate the work done here,
$W = - {P_{ext}} \times \Delta V\,$
Where $\,W = \,$Work
$\,{P_{ext}} = \,$External Pressure
$\,\Delta V = \,$Change in volume

Here, by converting the unit of volume and pressure we get;
 $\,\Delta V = 30.6l = 30.6 \times {10^{ - 3}}{m^3}\,$and $\,P = 1atm = 101325Pascal\,$
Therefore, $\,W = \, - 101325 \times 30.6 \times {10^{ - 3}}\,$
$\, = - 3100J\,$
Here, a negative sign shows that work is done by the system.
So, for this question option A is the correct answer.

Note:
Work done by a system in thermodynamics is the energy transmitted by the system to its surroundings by a process by which the system can exert macroscopic forces on its surroundings naturally, where those forces and their external effects can be calculated. So, in sign conventions, the energy transmitted here is represented as negative work done by the system on its surroundings.