SO2 (5.00 g) and CO2 (5.00 g) are placed in a 750.0 ml container at 50.0 °C. the partial pressure of SO2 in the container was 2.762 atm.
To calculate the partial pressure, we can calulate the total pressure
[tex]Total pressure = \frac{nRT}{V} SO2 + \frac{nRT}{V} CO2[/tex]
Where:
The amount of SO2 is n SO2 = mass/molar mass = m/M = [tex]\frac{5 }{32} + 16*2 = 0.078125[/tex] mol
The amount of CO2 is n CO2 = mass/molar mass = m/M = [tex]\frac{5}{14} + 16*2 = 0.1[/tex] mol
Total amount of gas is n(total) = [tex]n1 + n2 = 0.078 + 0.113 = 0.191[/tex] mol
T = 50+273 = 323 K
[tex]V = \frac{750}{1000} = 0.75 [/tex]liters
Total Pressure [tex]p*V = n*R*T[/tex]
Total Pressure [tex]= (0.078125*0.0821*\frac{323}{0.75} ) + (0.1*0.0821*\frac{323}{0.75}) = 2.7623 + 3.535 = 6.298[/tex] atm
Partial pressure = x SO2 * Total Pressure = [tex](\frac{no. of moles of SO2 }{total no. of moles} ) * Total pressure = (\frac{0.078125}{0.078125} +0.1) * 6.298 = 2.762 [/tex]atm
Grade: 9
Subject: chemistry
Chapter: pressure
Keywords: the container, the partial pressure, SO2, CO2, The mole fraction
