Answer:
\[{{p}_{1}}=\frac{40}{100}\times
\] Total pressure
\[=\frac{40}{100}\times
{{10}^{5}}=4\times {{10}^{4}}Pa\]
\[{{p}_{{{I}_{2}}}}=\frac{60}{100}\times
\] Total pressure
\[=\frac{60}{100}\times
{{10}^{5}}=6\times {{10}^{4}}Pa\]
\[{{K}_{p}}=\frac{{{({{p}_{1}})}^{2}}}{{{p}_{{{I}_{2}}}}}=\frac{(4\times
{{10}^{2}})}{6\times {{10}^{4}}}\]
\[=\frac{16}{6}\times
{{10}^{4}}Pa\]
\[=2.66\times {{10}^{4}}Pa\]
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