Solution:
Given,
![\[\begin{array}{*{35}{l}} Measure\text{ }of\text{ }the\text{ }gas\text{ },n~=\text{ }4.0\text{ }mol \\ ~ \\ Volume\text{ }of\text{ }the\text{ }gas~,V~=\text{ }5\text{ }dm3 \\ ~ \\ Tension\text{ }of\text{ }the\text{ }gas,p~=\text{ }3.32\text{ }bar \\ ~ \\ R\text{ }=\text{ }0.083\text{ }bar\text{ }dm3~K1~mol1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f082e941b5c8d1e8286454299ebf8cca_l3.png "Rendered by QuickLaTeX.com")
From the ideal gas condition, we get
![\[\begin{array}{*{35}{l}} pV\text{ }=\text{ }nRT \\ ~ \\ \Rightarrow ~T\text{ }=\text{ }pV~/~nR \\ \end{array}\begin{array}{*{35}{l}} ~ \\ =\text{ }\left( 3.32x5 \right)~/~\text{ }\left( 4x0.083 \right) \\ ~ \\ =\text{ }50K \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-150149258425b14afdb01953522c1ab6_l3.png "Rendered by QuickLaTeX.com")
Henceforth, the necessary temperature is 50 K
Work out the temperature of 4.0 mol of a gas involving 5 dm3 at 3.32 bar.
![\[\left( R\text{ }=\text{ }0.083\text{ }bar\text{ }dm3\text{ }K1\text{ }mol1 \right).\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0eea6131e45ed685db113cdc3bf5955a_l3.png "Rendered by QuickLaTeX.com")
Solution:
Given,
From the ideal gas condition, we get
Henceforth, the necessary temperature is 50 K