solution:
Warmth is free of the way under 2 conditions:
- At the point when the volume of the framework is kept steady
By first law of thermodynamics:
![\[\begin{array}{*{35}{l}} q\text{ }=\text{ }\Delta U\text{ }+\text{ }\left( -\text{ }w \right) \\ ~ \\ what's\text{ }more,\text{ }-\text{ }w\text{ }=\text{ }p\Delta V \\ ~ \\ Along\text{ }these\text{ }lines\text{ },\text{ }q\text{ }=\text{ }\Delta U\text{ }+\text{ }p\Delta V \\ ~ \\ \Delta V\text{ }=\text{ }0\text{ }\left( at\text{ }steady\text{ }volume \right) \\ ~ \\ Thus,\text{ }qv\text{ }=\text{ }\Delta U\text{ }+\text{ }0\text{ }=\text{ }\Delta U=\text{ }change\text{ }in\text{ }inner\text{ }energy \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ebf3225696083ae0af767ebf02be85b2_l3.png "Rendered by QuickLaTeX.com")
- At the point when the strain of the framework is kept consistent –
![\[\begin{array}{*{35}{l}} At\text{ }consistent\text{ }tension,\text{ }qp=\text{ }\Delta U\text{ }+\text{ }p\Delta V \\ ~ \\ However,\text{ },\text{ }\Delta U\text{ }+\text{ }p\Delta V\text{ }=\text{ }H \\ ~ \\ In\text{ }this\text{ }manner\text{ },\text{ }qp\text{ }=\text{ }H\text{ }=\text{ }change\text{ }in\text{ }enthalpy. \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3fa0fc205ec67224c2e37dae5c0fc484_l3.png "Rendered by QuickLaTeX.com")
Despite the fact that warmth is a way work yet warms consumed by the framework under certain particular conditions is free of way. What are those conditions? Clarify.
Despite the fact that warmth is a way work yet warms consumed by the framework under certain particular conditions is free of way. What are those conditions? Clarify.