91Ó°ÊÓ

The molar enthalpy of vaporization of water at \(373 \ K\) and \(1.00 \ atm\) is \(40.7 \ \frac{kJ}{mol}\).

The enthalpy change during vaporization, H, is the sum of the change in internal energy, U, and the work done by the system, W. In mathematical terms, this can be written as: \(H = U + W\)

Assuming the water vapor is an ideal gas, the work done against the atmosphere can be calculated with the formula: \(W = nRT\) Where n is the number of moles of the gas, R is the gas constant, and T is the absolute temperature. Here, we're not given the number of moles, but we can use a single mole since we're asked for a fraction of energy. So, n = 1 mol. The gas constant, R, in \(J/(mol K)\) is \(8.314 \frac{J}{molK}\). The temperature, T, is given as \(373 \ K\). Now, we can calculate the work done against the atmosphere: \(W = (1 \ mol) \times (8.314 \ \frac{J}{mol K}) \times (373 \ K)\) \(W = 3097.782 \ J\) Since we are given the enthalpy in kJ/mol, it's best to convert the work to kJ. \(W = 3.097782 \ kJ\)

We can now calculate the change in internal energy using the equation mentioned in step 2: \(U = H - W\) \(U = 40.7 \ kJ - 3.097782 \ kJ\) \(U = 37.602218 \ kJ\)

The total energy we're given is the enthalpy change, \(H = 40.7 \ kJ\). Now we can find the fraction of energy used for internal energy and work against the atmosphere: Fraction of energy used for internal energy: \(\frac{U}{H}\) \(\frac{37.602218 \ kJ}{40.7 \ kJ} \approx 0.924\) Fraction of energy used for work against the atmosphere: \(\frac{W}{H}\) \(\frac{3.097782 \ kJ}{40.7 \ kJ} \approx 0.076\) So, about \(92.4\%\) of the energy is used to change the internal energy of the water, and \(7.6\%\) is used to do work against the atmosphere.