Question

When 0.503 L of Ar at 1.20 atm and 227°C is mixed with 0.237 L of O₂ at 501 torr and 127°C in a 400. mL flask at 27°C, what is the pressure in the flask?

Answer 1

Ar

V=0.503L

P=1.20atm

T=227C=227+273=500K

R= 0.0821 L-atm/mole-k

PV=nRT

n=PV/RT = 1.20x0.503/0.0821x500 =0.0147 moles

number of moles of Ar= 0.0147 moles

O2

V= 0.237L

P=501torr =0.659 atm         [ 1 torr= 0.00131579 atm]

T=127C=127+273=400K

R= 0.0821 L-atm/mole-k

n=PV/RT = 0.659x0.237/0.0821x400 =0.00476 moles

number of moles of O2= 0.00476 moles

Total number of moles = n=0.0147+ 0.00476=0.01946 moles

n= 0.01946 moles

T=27C=27+273=300K

V=400ml = 0.4L

R= 0.0821 L-atm/mol-k

PV=nRT

P=nRT/V = 0.01946 x 0.0821x300/0.4=1.198atm

Pressure of the flask = 1.198 atm