
If the enthalpy change for the transition of liquid water of steam is \[30kJmo{{l}^{-1}}\] at \[{{27}^{o}}C\] the entropy change for the process would be (in \[Jmo{{l}^{-1}}{{K}^{-1}}\])
A.10
B.1
C.100
D.0%
Answer
414.6k+ views
Hint: Enthalpy is a thermodynamic system property that is defined as the sum of the system's internal energy plus the product of its pressure and volume. It is a state function that is used in many measurements in chemical, biological, and physical systems at constant pressure, which is provided conveniently by the large ambient atmosphere.
Complete answer:
The pressure-volume term expresses the effort required to establish the physical dimensions of the system, i.e. to make room for it by displacing its surroundings. Enthalpy, as a state function, is determined solely by the final configuration of internal energy, pressure, and volume, not by the path taken to get there.
In the International System of Units (SI), the joule is the unit of measurement for enthalpy. The calorie and the British thermal unit are two other historical conventional units that are still in use today (BTU).
The total enthalpy of a system cannot be directly measured because the internal energy contains components that are unknown, difficult to access, or irrelevant to thermodynamics. In actuality, a change in enthalpy is the preferred expression for measurements at constant pressure, because it simplifies the description of energy transfer.
In the above example,
\[{{H}_{2}}O(l)\to {{H}_{2}}O(g)\]
\[\vartriangle {{s}_{vapour}}=\dfrac{\vartriangle {{H}_{vapour}}}{T}\]
\[=\dfrac{30\times {{10}^{3}}}{300}\]
\[=100J/{{K}^{-1}}.mo{{l}^{-1}}\]
Thus, the enthalpy change for the transition of liquid water of steam is \[30kJmo{{l}^{-1}}\] at \[{{27}^{o}}C\] the entropy change for the process would be \[100J/{{K}^{-1}}.mo{{l}^{-1}}\]
Note:
An ideal gas's enthalpy is independent of its pressure and only depends on its temperature, which correlates to its internal energy. Real gases at common temperatures and pressures frequently approximate this behavior, making practical thermodynamic design and analysis easier.
Complete answer:
The pressure-volume term expresses the effort required to establish the physical dimensions of the system, i.e. to make room for it by displacing its surroundings. Enthalpy, as a state function, is determined solely by the final configuration of internal energy, pressure, and volume, not by the path taken to get there.
In the International System of Units (SI), the joule is the unit of measurement for enthalpy. The calorie and the British thermal unit are two other historical conventional units that are still in use today (BTU).
The total enthalpy of a system cannot be directly measured because the internal energy contains components that are unknown, difficult to access, or irrelevant to thermodynamics. In actuality, a change in enthalpy is the preferred expression for measurements at constant pressure, because it simplifies the description of energy transfer.
In the above example,
\[{{H}_{2}}O(l)\to {{H}_{2}}O(g)\]
\[\vartriangle {{s}_{vapour}}=\dfrac{\vartriangle {{H}_{vapour}}}{T}\]
\[=\dfrac{30\times {{10}^{3}}}{300}\]
\[=100J/{{K}^{-1}}.mo{{l}^{-1}}\]
Thus, the enthalpy change for the transition of liquid water of steam is \[30kJmo{{l}^{-1}}\] at \[{{27}^{o}}C\] the entropy change for the process would be \[100J/{{K}^{-1}}.mo{{l}^{-1}}\]
Note:
An ideal gas's enthalpy is independent of its pressure and only depends on its temperature, which correlates to its internal energy. Real gases at common temperatures and pressures frequently approximate this behavior, making practical thermodynamic design and analysis easier.
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