In: Chemistry
After 0.600 L of Ar at 1.50 atm and 248°C is mixed with 0.200 L of O2 at 333 torr and 123°C in a 400.−mLflask at 23°C, what is the pressure in the flask?
Ar
P = 1.5atm
V = 0.6L
T = 248+273 = 521K
PV = nRT
n = PV/RT
= 1.5*0.6/0.0821*521 = 0.021moles
O2
V = 0.2L
P = 333 torr = 333/760 = 0.438atm
T = 123 + 273 = 396K
PV = nRT
n = PV/RT
= 0.438*0.2/0.0821*396 = 0.00296mole
Total no of moles of mixture = 0.021+0.00296 = 0.02396moles
T = 23 + 273 = 296K
V = 400ml = 0.4L
PV = nRT
P = nRT/V
= 0.02396*0.0821*296/0.4 = 1.455atm >>>>>answer
the pressure in the flask = 1.455atm