Question

One mole of H₂O(g) at 1.00 atm and 100⁰ C occupies a volume of 30.6 L. When 1 mole of H₂O(g) is condensed to 1 mole of H₂O(l) at 1.00 atm and 100⁰C, 40.66 kJ of heat is released. If the density of H₂O(l) at this temperature and pressure is 0.996 g/cm³, calculate ΔE for the condensation of 1 mole of water at 1.00 atm and 100⁰C.

Short answer

The value of ΔE for the condensation of 1 mole of water at 1.00 atm and 100⁰C is -37.56kJ.

Step-by-step solution

Define Internal Energy

Internal energy is a thermodynamic property. It is basically the energy remained within the system.

Given data

$H_{2}O\left(g\right)\to H_{2}O\left(l\right)$

Let the work done by the system be denoted by the symbol w. The pressure is P. Initial volume is V₁ and final volume is V₂. The amount of heat transferred is q. The internal energy be$∆E$ .

The data given in the problem is as follows

$\begin{array}{rcl}q& =& -40.66kJ\\ V_1& =& 30.66L\end{array}$

Determine the work done

Volume of 1 mole of H₂O(l)

$$\begin{gathered} =1mol{H}_{2}O\left(l\right)\times \frac{18.02g}{mol}\times \frac{1c{m}^3}{0.996gm} \\ =18.1c{m}^3 \\ =18.1mL \end{gathered}$$

The work done for condensation will be as follows

$$\begin{gathered} w=-P∆V \\ =-P\left(V_2-V_1\right) \\ =-1atm\times \left(0.0181L-30.6L\right) \\ =30.6Latm \\ =30.6Latm\times \frac{101.3J}{Latm} \\ =3.1\times {10}^{3}J \\ =3.1kJ \end{gathered}$$

Now the internal energy will be

$$\begin{gathered} ∆E=q+w \\ ∆E=-40.66+3.10kJ \\ ∆E=-37.56kJ \end{gathered}$$

Therefore, the value of ΔE for the condensation of 1 mole of water at 1.00 atm and 100⁰C is -37.56kJ.