Question #fbcc4

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Question #fbcc4

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Answer 1 · 2017-05-15T22:48:40

P = 28 Atm (2 sig. figs.) but exact sig. figs. could be based upon the V = 500 ml data point => 1 sig. fig. => P = 30 Atm.

Explanation:

Using the Ideal Gas Law PV = nRT solve for P = nRT/P

n = $(moles) C O_{2}$ $("moles" ) CO_2$ (gas) = $\frac{25 g}{44 (\frac{g}{m o l})}$ $(25 g) / (44 (g/(mol))$ = $0.5681 mole$ $0.5681 "mole"$

$R = gas . constant = 0.08206 \frac{L-Atm}{mol-K}$ $R = "gas"."constant" = 0.08206 "L-Atm"/"mol-K"$

$T = T e m p e r a t u r e (Kelvin) = (25 + 273) K = 298 K$ $T = Temperature ("Kelvin") = (25 + 273)K = 298K$

$V = V o l u m e (L i t e r s) = (\frac{500}{1000}) L i t e r s = 0.500 Liters$ $V = Volume (Liters) = (500/1000) Liters = 0.500 "Liters"$

$P = \frac{(0.5681 mole) (0.08206 \frac{L-Atm}{mol-K}) (298 K)}{(0.500 L)}$ $P = [(0.5681 "mole")(0.08206 "L-Atm"/"mol-K")(298 K)] / [(0.500 L)]$

= $27.7885 A t m \Rightarrow 28 A t m (2 s i g . f i g s .) or ~ 30 A t m (1 s i g . f i g .)$ $27.7885 Atm => 28 Atm (2 sig. figs.) or ~ 30 Atm (1 sig. fig.)$

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Question #fbcc4

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