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Question: Assuming that, water vapor is an ideal gas, the internal energy change \((\Delta U)\)when \(1\) a mo...

Assuming that, water vapor is an ideal gas, the internal energy change (ΔU)(\Delta U)when 11 a mole of water is vaporized at 11 bar pressure and 1000C{100^0}C, (given: molar enthalpy of vaporization of water 41KJmol−141KJmo{l^{ - 1}} at 11 bar and 373K373K R=8.314JK−1mol−1R = 8.314J{K^{ - 1}}mo{l^{ - 1}}) will be:
A.4.100KJmol−14.100KJmo{l^{ - 1}}
B.3.7094KJmol−13.7094KJmo{l^{ - 1}}
C.37.904KJmol−137.904KJmo{l^{ - 1}}
D.41.00KJmol−141.00KJmo{l^{ - 1}}

Explanation

Solution

The enthalpy of vaporization of water is the amount of energy or heat required to which must be added to a liquid substance to transform a quantity of that substance into a gas. So, we will use the chemical equation of the above transformation.
Formula used:
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
Where, ΔH\Delta H is the molar enthalpy of vaporization of water, ΔU\Delta U is the change in internal energy, Δng\Delta {n_g} is the change in a number of moles of gaseous state, RR is the universal gas constant and TT is the temperature.

Complete step by step answer:
First, we will understand the basic meaning of Internal energy. In chemistry, the internal energy of a thermodynamic system is defined as the total energy of the closed system or in another way the sum of the potential energy of the system and the system’s kinetic energy. Now According to the question, we need to calculate internal energy for the reaction given below.
H2O(l)⇆H2O(g){H_2}O(l)\overset {} \leftrightarrows {H_2}O(g)
Now for the above change, we will calculate Δng\Delta {n_g}. We know that it is the change in the number of moles of water in a gaseous state in product and reactant. Therefore, Δng\Delta {n_g} will be,
Δng=1−0=1\Delta {n_g} = 1 - 0 = 1
Now for the given change we have H2O(l)⇆H2O(g),Δng=1,ΔH=41KJmol−1,R=8.314JK−1mol−1,T=373K{H_2}O(l)\overset {} \leftrightarrows {H_2}O(g),\Delta {n_g} = 1,\Delta H = 41KJmo{l^{ - 1}},R = 8.314J{K^{ - 1}}mo{l^{ - 1}},T = 373K. We can calculate the internal energy of the system for the above change using the formula or the relation of enthalpy and the internal energy of the system.
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
⇒ΔU=ΔH−ΔngRT\Rightarrow \Delta U = \Delta H - \Delta {n_g}RT
Now substituting the values we get,
⇒ΔU=41−1×8.314×373×10−3\Rightarrow \Delta U = 41 - 1 \times 8.314 \times 373 \times {10^{ - 3}}
⇒ΔU=37.904KJmol−1\Rightarrow \Delta U = 37.904KJmo{l^{ - 1}}
Therefore, the correct option is (C).

Note:
Internal energy of a system is important as the potential energies between molecules and atoms are important for understanding phase changes, chemical reactions, and nuclear reactions. For exothermic reactions the internal energy is negative.