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 = \[{{P}_{C{{H}_{4}}}}+{{P}_{C{{O}_{2}}}}\]
\[=5.543\times {{10}^{4}}Pa+2.771\times {{10}^{4}}Pa\]
\[=8.314\times
{{10}^{4}}Pa\]
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