If 33.0 L of a gas is collected at 23 °C and 255 kPa, what is its volume at STP?
First, let's use the Mendeleev-Clapeyron law to determine the quantity of matter of the gas: PV=nRT; n=(PV)/RT=(255*10^3*33*10^-3)/(296*8.31)=3.42 mol Each mole of the gas at standard conditions possesses a volume of 22.4 L, therefore; V=22.4*3,42=76.51 L
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