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What volume is occupied by 75.00 mol of argon gas at 25.0 C and 0.825 ATM?
Wiki User
∙ 12y ago
We use the ideal gas law here.
pressure * volume = moles * R * temperature in Kelvin ( 25.0 C + 273.15 = 298.15 K )
PV = nRT
(0.825 atm)(volume) = (75.00 moles Ar)(0.08206 L*atm/mol*K)(298.15 K)
Volume = (75.00 * 0.08206 * 298.15)/(0.825
Volume = 1834.964/0.825
= 2224.199 Liters
which is for sigi figis sake
2.22 X 103
--------------------- Liters in volume
Wiki User
∙ 12y ago
Wiki User
∙ 10y agoFirst, convert torrs (mmHg) to atmospheres: 1atm=760torr, so 8.25torr=0.011atm. That's about the atmospheric pressure on the planet Mars. Now, convert your 27 Celsius to kelvins: 27+273=300. Now, plug it all in to the ideal gas law equation, using 0.08 for the value of "R" the ideal gas constant: PV=nRT. 0.011(V)=4(0.08)300; solving for V gets you about 8,700 liters. Intuitively, this makes sense, because at very low pressure, your volume will be relatively large.
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