22 , g of dry ice is placed in a 600 , mL emp | vedclass

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(A) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,Given: $V = 600 \, mL = 0.6 \, L$,$m = 22 \, g$,$M = 44 \, g/mol$,$R = 0.0821 \, L \cdot a

Vedclass · Accepted Answer

$20.4$

Vedclass · Answer

(A) Using the ideal gas equation $PV = nRT = \frac{m}{M}RT$,Given: $V = 600 \, mL = 0.6 \, L$,$m = 22 \, g$,$M = 44 \, g/mol$,$R = 0.0821 \, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,$T = 25 + 273 = 298 \, K$.Calculating the pressure exerted by $CO_2$ gas:$P = \frac{m 	imes R 	imes T}{M 	imes V} = \frac{22 	imes 0.0821 	imes 298}{44 	imes 0.6} = 0.5 	imes \frac{24.4658}{0.6} = 20.388 \approx 20.4 \, atm$.Since the container was initially empty (vacuum) or the question implies the total pressure inside the sealed vessel after evaporation,the pressure is $20.4 \, atm$.

$22 \, g$ of dry ice is placed in a $600 \, mL$ empty container at $25 \, ^\circ C$ and sealed tightly. If all the $CO_2$ evaporates,what will be the final pressure inside the container in $atm$?

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