At 27,^oC temperature,4,mol Cl_2,4,mol N_2,an | vedclass

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Question

(A) The total number of moles of the gas mixture is $n_{total} = n_{Cl_2} + n_{N_2} + n_{O_2} = 4 + 4 + 2 = 10 \,mol$. The temperature in Kelvin is $T

Vedclass · Accepted Answer

$49.88$

Vedclass · Answer

(A) The total number of moles of the gas mixture is $n_{total} = n_{Cl_2} + n_{N_2} + n_{O_2} = 4 + 4 + 2 = 10 \,mol$. The temperature in Kelvin is $T = 27 + 273 = 300 \,K$. The volume of the vessel is $V = 5 \,L$. Using the ideal gas equation $PV = nRT$,the total pressure $P$ is given by $P = \frac{n_{total}RT}{V}$. Substituting the values: $P = \frac{10 \,mol 	imes 8.34 	imes 10^{-2} \,bar \,L \,mol^{-1} \,K^{-1} 	imes 300 \,K}{5 \,L}$. $P = \frac{10 	imes 0.0834 	imes 300}{5} = \frac{250.2}{5} = 50.04 \,bar$. Given the options,the closest value is $49.88 \,bar$.

At $27\,^oC$ temperature,$4\,mol$ $Cl_2$,$4\,mol$ $N_2$,and $2\,mol$ $O_2$ are filled in a $5\,L$ closed vessel. Find the total pressure of the gaseous mixture. $(R = 8.34 \times 10^{-2} \,bar \,L \,mol^{-1} \,K^{-1})$ (in $,bar$)

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