Question #e8863

2 Answers
Nov 19, 2015

The volume of 0.00988 moles of Oxygen @ STP = 224.5 ml

Explanation:

Since the condition is at STP, therefore the value of P = 0.98692 atm, T= 273.15 K and the R = 0.0821 li- atm/mole-K. The no. of moles of Oxygen is given as 0.00988 moles.

Using the Ideal Gas Law:

PV = nRT , divide both sides by P to isolate the V
V = nRT/P, the desired formula, then substitute the given data
V = {0.00988cancel(mol)}{0.0821 li-cancel(atm)/(cancel(mol)-cancel(K)}{273.15cancel(K)}/{0.98692cancel(atm)}
V = 0.2215519 li/0.98692
V = 0.2244882 li, convert to ml
V = 224.5 ml, the volume of 0.00988 moles of Oxygen at STP

Nov 20, 2015

The volume of oxygen gas is "224 mL".

Explanation:

"STP" is "273.15 K" and "100 kPa".

Use the ideal gas law to answer this question. The equation for the ideal gas law is "PV=nRT".

Given/Known
P="100 kPa"
T="273.15 K"
n="0.00988 mol"
R="8.3144598 L kPa K"^(-1) "mol"^(-1)"

Unknown
Volume in mL

Solution
Rearrange the equation to isolate V and solve. Then multiply the volume in liters times 1000 to get mL.

V=(nRT)/(P)

V=((0.00988cancel"mol")xx(8.3144598"L" cancel"kPa" cancel("K"^(-1)) cancel("mol"^(-1)))xx(273.15cancel"K"))/(100cancel"kPa")="0.224 L"

Convert 0.224 L to mL

0.224"L"xx(1000"mL")/(1"L")="224 mL"