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What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 27 °C ?

sudhanshu , 12 Years ago
Grade 12
anser 4 Answers
Gaurav
Hello Student
Molar mass of Mathane = 16 g
Moles = mass / molar mass
= 3.2 g / 16 g/mol
= 0.20 mol
for carbon dioxide (molar mass = 44 g/mol)
moles = mass / molar mass
= 4.4 g / 44 g/mol
= 0.10 mol

Total number of moles = 0.30 mol of gas

Now
PV = nRT
so P = nRT / V
n = 0.30 mol
R = 8.31 kPa.L/mol.K
T = 27°C = 300 K
V = 9.0 L
P = (0.30 mol)(8.31 kPa.L/mol.K)(300 K) / (9.0 L)
= 83 kPa
Last Activity: 10 Years ago
Qunoot usama
Molar mass of CH4: 16
No of moles of CH4 =Given wt/molecular wt
                                    =3.2/16
                                   =0.2 moles
Molar mass of CO2 :44
No of moles of CO2 =4.4/44
                                   = 0.1
Total no of moles=0.1+0.2=0.3
Now,PV=nRT
          P=nRT/V
             = 0.3×0.082×300/9
            =0.1×0.082×100
           = 8200× 10^-4
          = 0.82 atm
no of moles of CO2=44/4.4
  
Last Activity: 7 Years ago
Rohit Upadhyay
P=n/V×RT=w/M×RT/V.
Pch4=(3.2/16 mol)0.821 dm^3 atmk^-1 mol^-1×300k=0.55 atm
Similarly, Pco2 =4.4/44mol×0.0821 dm^3 atm k^-1 mol^-1×300K = 0.27 atm
Ptotal=0.55+0.27 = 0.82 atm
Last Activity: 7 Years ago
Kushagra Madhukar
Dear student,
Please find the attached solution to your problem.
Molar mass of Mathane = 16 g
No. of Moles of Methane = 3.2 g / 16 g/mol = 0.20 mol
Molar mass of CO2 = 44 g/mol
No. of moles of CO2 = 4.4 g / 44 g/mol = 0.10 mol
 
Hence,Total number of gaseous moles = 0.30 mol
Now,
PV = nRT
so, P = nRT / V
n = 0.30 mol
R = 8.31 kPa.L/mol.K
T = 27°C = 300 K
V = 9.0 L
P = (0.30 mol)(8.31 kPa.L/mol.K)(300 K) / (9.0 L) = 83 kPa = 0.82 atm
 
Hope it helps.
Thanks and regards,
Kushagra
 
Last Activity: 5 Years ago
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