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Question: Calculate w and \(\Delta U\) for the conversion of 0.5 moles of water at \({{100}^{\circ }}C\) to st...

Calculate w and ΔU\Delta U for the conversion of 0.5 moles of water at 100∘C{{100}^{\circ }}C to steam at 1 atm pressure. The heat of vaporization of water at 100∘C{{100}^{\circ }}C is 40670 J/mol.

Explanation

Solution

The work done by the system can be calculated by multiplying the pressure into the change in volume. The change in internal energy is equal to the sum of heat and work are done by the system.

Complete step by step answer:
The volume of the steam can be calculated by the ideal gas equation:
PV=nRTPV=nRT
The volume will be: V=nRTPV=\dfrac{nRT}{P}, where n is the number of moles, P is the pressure, T is the temperature, and R is the universal gas constant.
The number of moles is 0.5 and the pressure given is 1 atm.
Since, the pressure is given in atm, the value of R will be 0.0821 liter atmosphere degreee−1 mol−10.0821\text{ liter atmosphere degree}{{\text{e}}^{-1}}\text{ mo}{{\text{l}}^{-1}}
So, putting all the value, we get
V=nRTP=0.5 x 0.0821 x 3731=15.3LV=\dfrac{nRT}{P}=\dfrac{\text{0}\text{.5 x 0}\text{.0821 x 373}}{1}=15.3L
So, the volume of steam is 15.3L and the volume of water is negligible.
Hence, the change in volume = volume of steam – volume of water
= 15.3 - negligible = 15.3L=\text{ 15}\text{.3 - negligible = 15}\text{.3L}
The work done by the system is calculated by multiplying the pressure to the change in volume.
w=Pext x ΔVw={{P}_{ext}}\text{ x }\Delta \text{V}
So, the work done will be,
w= 1 x 15.3 = 15.3 liter-atmw=\text{ 1 x 15}\text{.3 = 15}\text{.3 liter-atm}
To convert the work done to joules multiply the value with 101.3.
w= 15.3 x 101.3 = 1519.89Jw=\text{ 15}\text{.3 x 101}\text{.3 = 1519}\text{.89J}
The work done should be negative because the work is done by the system on the surrounding.
w=−1549.89Jw=-1549.89J
The heat of vaporization of water at 100∘C{{100}^{\circ }}C is 40670 J/mol.
So, the heat required to convert 0.5 mole of water to steam will be:
q=n x ΔHvapq=n\text{ x }\Delta {{\text{H}}_{vap}}
q=0.5 x 40670 = 20335 Jq=0.5\text{ x 40670 = 20335 J}
Now, according to the first law of thermodynamics we know that,
ΔU=q+w\Delta U=q+w
ΔU=20335−1549.89= 18785.11 J\Delta U=20335-1549.89=\text{ 18785}\text{.11 J}

Hence, the w=−1549.89Jw=-1549.89J, and ΔU= 18785.11 J\Delta U=\text{ 18785}\text{.11 J}.

Note: According to the sign conventions, w is taken positive if the work is done in the system and w is negative if the work is done by the system. Since the heat of vapourization is given joules, the work done is also converted into joules. The temperature should always be taken in K. The Kelvin temperature is calculated by- 273+ ∘C273+{{\text{ }}^{\circ }}C.