In a container,the mole ratio of two gases A | vedclass

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(C) The partial pressure of a gas is given by the product of its mole fraction and the total pressure $(P_i = x_i 	imes P_{total})$.Given the mole ra

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$2.5 	imes 10^5 \ Pa$

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(C) The partial pressure of a gas is given by the product of its mole fraction and the total pressure $(P_i = x_i 	imes P_{total})$.Given the mole ratio $n_A : n_B = 2 : 5$,the total number of parts is $2 + 5 = 7$.The mole fraction of gas $B$ is $x_B = \frac{5}{7}$.The total pressure $P_{total} = 3.5 	imes 10^5 \ Pa$.Therefore,the partial pressure of $B$ is $P_B = \frac{5}{7} 	imes 3.5 	imes 10^5 \ Pa$.$P_B = 5 	imes 0.5 	imes 10^5 \ Pa = 2.5 	imes 10^5 \ Pa$.

In a container,the mole ratio of two gases $A$ and $B$ is $2 : 5$. The total pressure of the container is $3.5 \times 10^5 \ Pa$. If all molecules of $A$ are removed from the container,what will be the pressure exerted by the molecules of $B$?

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