Question

What is the work done against the atmosphere when 25 grams of water evaporates at 73 K against a constant external pressure of 1 atm? Assume that stream obeys perfect gas bove laws. Given that the molar enthalpy of vaporisation is 9.72 Kcal/mol. What is the change of internal energy in the above process?
(A) 1294 cal, 11247 cal
(B) 921.4 cal, 11074 cal
(C) -1029.1 cal, 12470 cal
(D) 1129.3 cal, 10207 cal

Answer 1

Work is defined as force multiplied by distance in physical sciences such as physics and chemistry. If there is movement in the force's direction, work is created. The joule is the SI unit of work (J). This is the work done by a force of one newton (N) over a one-meter displacement (m).

Complete answer:
An ideal gas is a theoretical gas made up of a large number of randomly moving point particles with no interparticle interactions. Because it obeys the ideal gas law, a simplified equation of state, and is susceptible to statistical mechanics analysis, the ideal gas notion is helpful. If the interaction is fully elastic or viewed as point-like collisions, the criterion of zero interaction can frequently be waived.
We know that PV = nRT
Now the Volume of 25g of water at 373K and 1 atm is given as
${{\mathbf{V}}_1} = {\mathbf{0}}.{\mathbf{018L}} \times \dfrac{{{\mathbf{25}}}}{{{\mathbf{18}}}}$
Then using ideal gas equation Volume of 25g of water in vapour state at 373K
${{\text{V}}_2} = \dfrac{{{\text{nRT}}}}{{\text{P}}} = \dfrac{{25 \times 0.0821 \times 373}}{{18 \times 1}} = 42.53\;{\text{L}}$
Now work done,
${\mathbf{W}} = - {\mathbf{P}}\left( {{{\mathbf{V}}_2} - {{\mathbf{V}}_1}} \right) = - {\mathbf{1}}({\mathbf{42}}.{\mathbf{53}} - {\mathbf{0}}.{\mathbf{025}})Latm$
=-42.505 Latm
=-4305.7 J
Convert J to cal, we get
=-1029.1 cal
Now the Change in internal energy is given as $= \Delta {{\text{H}}_{{\text{vap}}}} + {\text{W}}$
$= 9.27 \times {10^3}{\text{cal}} \times \dfrac{{25}}{{18}} - 1029.1$
=12470.9 cal
Hence option C is correct.

Note:
A thermodynamic system's enthalpy is defined as the sum of the system's internal energy and the product of its pressure and volume. It is a state function that is employed in many chemical, biological, and physical tests under constant pressure, which is easily given by the vast ambient environment. The effort necessary to create the system's physical dimensions, i.e. to make space for it by displacing its surrounds, is expressed by the pressure–volume term.