question_answer 1)

What mil be the pressure exerted (in pascal) by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a\[9d{{m}^{3}}\]flask at\[27{{\,}^{{}^\circ }}C\]?

Answer:

\[PV=\frac{w}{W}RT\] Pressure of methane: \[w=3.2g,\]\[M=16g\,mo{{l}^{-1}}\] \[T=300K\] \[V=9\times {{10}^{-3}}{{m}^{3}},\]\[R=8.314Pa\,{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}\] Putting these values in Eqn. (i), we get \[{{P}_{C{{H}_{4}}}}=\frac{3.2}{16}\times \frac{8.314\times 300}{9\times {{10}^{-3}}}=5.543\times {{10}^{4}}Pa\] Pressure of\[\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}\]: \[w=4.4g,\] \[M=44g\,mo{{l}^{-1}}\]\[T=300K\] \[V=9\times {{10}^{-3}}{{m}^{3}},\]\[R=8.314Pa\,{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}\] Putting these values in equation (i) we get \[{{P}_{C{{O}_{2}}}}=\frac{4.4}{44}\times \frac{8.314\times 300}{9\times {{10}^{-3}}}\] \[=2.771\times {{10}^{4}}Pa\] Total pressure of mixture = \[{{P}_{C{{H}_{4}}}}+{{P}_{C{{O}_{2}}}}\] \[=5.543\times {{10}^{4}}Pa+2.771\times {{10}^{4}}Pa\] \[=8.314\times {{10}^{4}}Pa\]