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The change in internal energy \( (U) \) can be brought about in two ways:
(i) Either by allowing the heat to flow into the system or out of the system
(ii) By doing work on the system or the work done by the system
Using the symbol \( q \) to represent heat transferred to system and using work done by the system \( -w \), we can represent the internal energy change of a system, \( \Delta U \), as :
\[
q=\Delta U+(-w)
\]
(First law of thermodynamics)
If the reaction is carried out in a closed container with constant volume, so that \( \Delta V=0 \)
Hence,
\[
q_{v}=\Delta U
\]
On the other hand, if a reaction is carried out in open vessel that keeps the pressure constant and allows the volume of the system to change freely. In such case, \( \Delta V \neq 0 \) and \( -w=P \cdot \Delta V \).
\( \begin{array}{ll}\text { Hence, } & q_{p}=\Delta U+P \Delta V \\ \text { Also, } & q_{p}=q_{v}+\Delta n_{g} R T\end{array} \)
As reactions carried out at constant pressure are so common, the heat change for such process is given a special symbol \( \Delta H \), called the enthalpy change of the reaction. The enthalpy \( (H) \) of the system is the name given to the quantity \( (U+P V) \).
A system is provided \( 50 \mathrm{~kJ} \) energy and work done on the system is \( 100 \mathrm{~J} \). The change in internal energy is:
(A) \( 150 \mathrm{~J} \)
(B) \( 50 \mathrm{~kJ} \)
(C) \( 50.1 \mathrm{~kJ} \)
(D) \( 49.9 \mathrm{~kJ} \)
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