Question

after 0.6 L of Ar at 1.46 atm and 242 C is mixed with 0.2 L of O2 at 360 torr and 147 C in a 400 mL flask at 21 C, what is the pressure in the flask?

Answer 1

Ar

PV = nRT

n = PV/RT

T = 242C = 242 +273 = 515K

n = 1.46*0.6/0.0821*515 = 0.876/42. 28   =0.0207 moles

PV = nRT

n = PV/RT

T = 147C = 147 + 273 = 420K

P = 360 torr = 360 /760 = 0.473 atm

n = PV/RT

   = 0.473*0.2/0.0821*420 = 0.0946/34.482 = 0.00274 moles

in Flask

V = 400ml = 0.4 L

T = 21C = 21+ 273 = 294K

total no of moles in the flask = nAr + nO2 = 0.0207 + 0.00274 = 0.02344 moles

Pv = nRT

P = nRT/V

= 0.02344*0.0821*294/0.4 = 1.415 atm

pressure in the flask = 1.415 atm