At 200 K,an ideal gas (X) present in a 1 L | vedclass

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(A) Initial concentration of gas $(X)$ is $1 \ mol \ L^{-1}$ in a $1 \ L$ flask,so initial moles $(n_1) = 1 \ mol$.Using the ideal gas equation $PV =

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$18.04$

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(A) Initial concentration of gas $(X)$ is $1 \ mol \ L^{-1}$ in a $1 \ L$ flask,so initial moles $(n_1) = 1 \ mol$.Using the ideal gas equation $PV = nRT$,the initial pressure $(P_1)$ is:$P_1 = \frac{n_1 RT}{V} = \frac{1 	imes 0.082 	imes 200}{1} = 16.4 \ atm$.After adding $0.1 \ mol$ of $X$,the new number of moles $(n_2) = 1 + 0.1 = 1.1 \ mol$.Since temperature $(T)$ and volume $(V)$ remain constant,the pressure is directly proportional to the number of moles $(P \propto n)$.Therefore,$\frac{P_1}{n_1} = \frac{P_2}{n_2}$.$P_2 = P_1 	imes \frac{n_2}{n_1} = 16.4 	imes \frac{1.1}{1} = 18.04 \ atm$.

At $200 \ K$,an ideal gas $(X)$ present in a $1 \ L$ flask has a concentration of $1 \ mol \ L^{-1}$. At the same temperature,$0.1 \ mol$ of $X$ is added into the vessel. What is the final pressure of the gas in $atm$?
(Given $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)

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At $200 \ K$,an ideal gas $(X)$ present in a $1 \ L$ flask has a concentration of $1 \ mol \ L^{-1}$. At the same temperature,$0.1 \ mol$ of $X$ is added into the vessel. What is the final pressure of the gas in $atm$?(Given $R = 0.082 \ L \ atm \ mol^{-1} \ K^{-1}$)