Question

Assuming that, water vapor is an ideal gas, the internal energy change $\Delta U$ when $1mole$ of water is vaporized at 1 bar pressure and $100^\circ C$ , (given: molar enthalpy of vaporization of water at 1 bar and $373K$ = $41kJmo{l^{ - 1}}$ and $R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$ ) will be:
A.$41.00Jmo{l^{ - 1}}$
B.$4.100Jmo{l^{ - 1}}$
C.$3.7904Jmo{l^{ - 1}}$
D.$37.904Jmo{l^{ - 1}}$

Answer 1

We need to study the ideal gas equation and accordingly study the incorporation of internal energy and enthalpy in the ideal gas equation. An ideal gas is one whose particles have negligible volume, are equally sized and do not possess intermolecular forces, have a random motion and perfect elastic collision with no energy loss. The ideal gas equation is given as $PV = nRT$ .

Complete step by step answer:
As we know that an ideal gas strictly obeys the equation $PV = nRT$ where Pressure ($P$), Volume $\left( V \right)$, number of moles of gas $\left( n \right)$ and Temperature $\left( T \right)$ are taken into consideration. According to the definition of enthalpy ($H$) which states that enthalpy is the sum of the internal energy ($U$) and the product of the pressure and volume of a thermodynamic system. The equation is given as
$H = {\text{U}} + PV$
But for an ideal gas $PV = nRT$. Therefore,
$H = {\text{U}} + nRT\;$
We now apply the above equation in the given question:
Given,
$\Delta H = 41kJmo{l^{ - 1}}$
$R = 8.314{\text{ }}J{K^{ - 1}}mo{l^{ - 1}}$
$T = 373K$
$n = 1mol$
We are to find the internal energy change $\Delta U$ using the equation $H = {\text{U}} + nRT$
Or $\Delta H = \Delta {\text{U}} + nRT$
Now we can substitute the known values we get,
Or $41kJmo{l^{ - 1}} = \Delta U + 8.314J{K^{ - 1}}mo{l^1} \times 373K$
On simplification we get,
$\Delta U = 37904 kJmo{l^{ - 1}}$
To convert it to joules per mole, we divide $37904kJmo{l^{ - 1}}$ by $1000$
On calculating, $\Delta U = 37.904{\text{ }}Jmo{l^{ - 1}}$
Therefore, the option D is correct.

Note:
We must note that the most important step in these types of problems is the calculation of the number of moles $\left( n \right)$ which is to be used in the ideal gas equation. For example, according to the above question, the vaporization of water will have the reaction $2{H_2}O(l) \to 2{H_2}(g) + {O_2}(l)$ .Since we are calculating the internal energy change, we calculate the change in number of moles $\Delta n$=${n_{products}} - {n_{reac\tan ts}}$= $3 - 2 = 1$ . Hence we take $n$ as one mole.