# Question fbcc4

May 15, 2017

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 = $\left(\text{moles}\right) C {O}_{2}$(gas) = (25 g) / (44 (g/(mol)) = $0.5681 \text{mole}$

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

$T = T e m p e r a t u r e \left(\text{Kelvin}\right) = \left(25 + 273\right) K = 298 K$

$V = V o l u m e \left(L i t e r s\right) = \left(\frac{500}{1000}\right) L i t e r s = 0.500 \text{Liters}$

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

= 27.7885 Atm => 28 Atm (2 sig. figs.) or ~ 30 Atm (1 sig. fig.)#