A sample containing 5.20g O2 gas has a volume of 19.0L, pressure and temperature remain constant, Oxygen is released until the volume is 7.00L how many moles of O2 are removed
I know i convert 5.20g to moles which give me 0.1625 moles of O2 and i use the P=nRT/V formula but i still can't figure out the answer
To find the number of moles of oxygen gas removed, you can use the Ideal Gas Law equation, which is:
PV = nRT
where:
P is the pressure (constant)
V is the volume (initial and final)
n is the number of moles (unknown)
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature (constant)
Since the pressure and temperature are constant, we can simplify the equation to:
P1V1 = nRT
Now let's plug in the given values:
P1 = constant (not given)
V1 = 19.0 L (initial volume)
n = 0.1625 moles (initial amount of O2, calculated from the mass)
R = 0.0821 L·atm/mol·K (constant)
T = constant (not given)
So, the equation becomes:
P1 * 19.0 = 0.1625 * R * T
Next, let's find the final number of moles of oxygen gas (n2) after the volume decreases to 7.00 L. We'll use the same equation, but with the final volume (V2):
P1 * V2 = n2 * R * T
Plugging in the given values:
P1 * 7.00 = n2 * R * T
Now divide both equations to cancel out the constants (P1, R, and T):
(P1 * 19.0) / (P1 * 7.00) = (0.1625 * R * T) / (n2 * R * T)
19.0 / 7.00 = 0.1625 / n2
Cross-multiplying:
19.0 * n2 = 7.00 * 0.1625
n2 = (7.00 * 0.1625) / 19.0
Calculate n2:
n2 = 0.8382 moles
Therefore, approximately 0.8382 moles of O2 gas are removed when the volume is reduced from 19.0 L to 7.00 L while keeping the pressure and temperature constant.