How many moles of O2 contained in 5.25L of a gas at 26 degrees C and 1.2 atm?
Since PV=nRT, we have
PV/nT = R is constant.
So, use STP and your given numbers:
(1*1)/(1*273) = (1.2*5.25)/((273+26)n)
Just solve for n.
I think there is an extra 273 (on the left side) in this solution that doesn't belong there.
PV = nRT
n = PV/RT
n = 1.2*5.25/(0.08206*299)
n = 0.26
Oh, yeah: 22.4L at STP, not 1L.
(1*22.4)/(273*1) = (1.2*5.25)/(299n)
n = 0.256
My 273 is included in the .08026 factor you inserted.
right. and I tried to get 0.08206 out of 1/273 but no matter how hard I tried it wouldn't work. But 22.4/273 = 0.08205 or
22.404/273 = 0.08206 which is the number usually quoted. I expect a mathematician will tell me that 22.404 is a lot of s.f. that probably don't belong there.
To find the number of moles of O2 in a gas, you can use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, convert the given temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15
T(K) = 26 + 273.15 = 299.15 K
Next, rearrange the Ideal Gas Law equation to solve for the number of moles (n):
n = PV / RT
Now, substitute the given values into the equation:
P = 1.2 atm
V = 5.25 L
R = 0.0821 L·atm/(mol·K) (gas constant for atm)
T = 299.15 K
n = (1.2 atm * 5.25 L) / (0.0821 L·atm/(mol·K) * 299.15 K)
Performing the calculation:
n = 7.78 moles
Therefore, there are approximately 7.78 moles of O2 in the given gas.