A 20 $g$ chunk of dry ice is placed in an empty 0.75 $L$ wire bottle tightly closed what would be the final pressure in the bottle after all $C O_{2}$ has been evaporated and temperature reaches to 25° $C$ ?

15.8

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21.2

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12.5

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Solution

The correct option is A 15.8
$w$ = 20 $g$ dry $C O_{2}$
which will evaporate to develop pressure

$m$ = 44, $V$ = 0.75 $L$ , $P$ = ?, T = 298 $K$

$P V = \frac{w}{m} R T$ (as $P V = n R T$ & $n = \frac{w}{m}$ )

$P \times 0.75 = \frac{20}{44} 20 / 445 0.0821 \times 298$

P = 14.228 $a t m$

Pressure inside the bottle = $P$ + $a t m$ pressure
$= 14.828 + 1 = 15.828$ $a t m$

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A 20 g chunk of dry ice is placed in an empty 0.75 L wire bottle tightly closed what would be the final pressure in the bottle after all CO2 has been evaporated and temperature reaches to 25°C?

The correct option is A 15.8w = 20 g dry CO2which will evaporate to develop pressure m = 44, V = 0.75 L, P = ?, T = 298 K PV=wmRT (as PV=nRT & n=wm)P×0.75=204420/4450.0821×298P = 14.228 atm Pressure inside the bottle = P + atm pressure =14.828+1=15.828 atm

A 20 $g$ chunk of dry ice is placed in an empty 0.75 $L$ wire bottle tightly closed what would be the final pressure in the bottle after all $C O_{2}$ has been evaporated and temperature reaches to 25° $C$ ?