A and B are ideal gases. At T(K),2 L of A wi | vedclass

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(A) For an ideal gas,the product of pressure and volume $(PV)$ is proportional to the number of moles $(n)$ at a constant temperature $(T)$.Using the

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$2$

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(A) For an ideal gas,the product of pressure and volume $(PV)$ is proportional to the number of moles $(n)$ at a constant temperature $(T)$.Using the principle of conservation of moles,the total pressure of the mixture $(P_{mix})$ in the final volume $(V_{final} = 100 \ L)$ is given by the sum of the partial pressures of $A$ and $B$.Initial state of $A$: $P_1 = 1 \ bar$,$V_1 = 2 \ L$.Initial state of $B$: $P_2 = p_B \ bar$,$V_2 = 4 \ L$.Final state of mixture: $P_{mix} = 0.1 \ bar$,$V_{final} = 100 \ L$.According to Boyle's Law $(P_1V_1 + P_2V_2 = P_{mix}V_{final})$:$(1 \ bar 	imes 2 \ L) + (p_B \ bar 	imes 4 \ L) = 0.1 \ bar 	imes 100 \ L$$2 + 4p_B = 10$$4p_B = 10 - 2$$4p_B = 8$$p_B = 2 \ bar$.Therefore,the value of $p_B$ is $2$.

$A$ and $B$ are ideal gases. At $T(K)$,$2 \ L$ of $A$ with a pressure of $1 \ bar$ is mixed with $4 \ L$ of $B$ with a pressure $p_B \ bar$ in a $100 \ L$ flask. The pressure exerted by the gaseous mixture is $0.1 \ bar$. What is the value of $p_B$ in $bar$?

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