Question

Consider a gas mixture in a 2.00-dm3 flask at 27.0 ºC. For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the mixture in mole percent. a) 1.00 g H2 and 1.00 g O2 b) 1.00 g N2 and 1.00 g O2 c) 1.00 g CH4 and 1.00 g NH3

Answer 1

Explanation:

a)

Moles of hydrogen gas = $n_1=\frac{1.00 g}{2 g/mol}=0.5 mol$

Moles of oxygen gas = $n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol$

Total moles in container = $n=n_1+n_2=0.5 mol+0.03125 mol=0.53125 mol$

Total pressure of mixture = P

Temperature of the mixture = $T = 27^oC =27+273K= 300 K$

Volume of the container in which mixture is kept = $2.00 dm^3 =2.00 L$

$1 dm^3=1 L$

$P=\frac{nRT}{V}$ (from ideal gas equation )

$P=\frac{0.53125 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=6.54 atm$

Partial pressure of the hydrogen gas :

= $p_1=P\times \chi_1=P\times \frac{n_1}{n}$

$=6.54 atm\times \frac{0.5 mol}{0.53125 mol}=6.16 atm$

Partial pressure of the oxygen gas :

= $p_2=P\times \chi_2=P\times \frac{n_2}{n}$

$=6.54 atm\times \frac{0.03125 mol}{0.53125 mol}=0.38 atm$

hydrogen

= $\frac{n_1}{n}\times 100=\frac{0.5 mol}{0.53125 }\times 100$

= 94.12%

oxygen :

= $\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.53125  mol}\times 100$

= 5.88%

b)

Moles of nitrogen gas = $n_1=\frac{1.00 g}{28 g/mol}=0.03571 mol$

Moles of oxygen gas = $n_2=\frac{1.00 g}{32 g/mol}=0.03125 mol$

Total moles in container = $n=n_1+n_2=0.03571 mol+0.03125 mol=0.06696 mol$

Total pressure of mixture = P

Temperature of the mixture = $T = 27^oC =27+273K= 300 K$

Volume of the container in which mixture is kept = $2.00 dm^3 =2.00 L$

$1 dm^3=1 L$

$P=\frac{nRT}{V}$ (from ideal gas equation )

$P=\frac{0.06696 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=0.82 atm$

Partial pressure of the nitrogen gas :

= $p_1=P\times \chi_1=P\times \frac{n_1}{n}$

$=0.82 atm\times \frac{0.03571 mol}{0.06696 mol}=0.44 atm$

Partial pressure of the oxygen gas :

= $p_2=P\times \chi_2=P\times \frac{n_2}{n}$

$=0.82 atm\times \frac{0.03125 mol}{0.06696 mol}=0.38 atm$

Composition of each in mole percent :

nitrogen

= $\frac{n_1}{n}\times 100=\frac{0.03571 mol}{0.06696 }\times 100$

= 53.33%

oxygen :

= $\frac{n_2}{n}\times 100=\frac{0.03125 mol}{0.06696 mol}\times 100$

= 46.67%

c)

Moles of methane gas = $n_1=\frac{1.00 g}{16 g/mol}=0.0625 mol$

Moles of ammonia gas = $n_2=\frac{1.00 g}{17g/mol}=0.0588 mol$

Total moles in container = $n=n_1+n_2=0.0625 mol+0.0588 mol=0.1213 mol$

Total pressure of mixture = P

Temperature of the mixture = $T = 27^oC =27+273K= 300 K$

Volume of the container in which mixture is kept = $2.00 dm^3 =2.00 L$

$1 dm^3=1 L$

$P=\frac{nRT}{V}$ (from ideal gas equation )

$P=\frac{0.1213 mol\times 0.0821 atm  L/mol K\times 300 K}{2.00 L}=1.49 atm$

Partial pressure of the methane gas :

= $p_1=P\times \chi_1=P\times \frac{n_1}{n}$

$=1.49 atm\times \frac{0.0625 mol}{0.1213mol}=0.77 atm$

Partial pressure of the ammonia gas :

= $p_2=P\times \chi_2=P\times \frac{n_2}{n}$

$=1.49 atm\times \frac{0.0588 mol}{0.1213mol}=0.72 atm$

Composition of each in mole percent :

Methane :

= $\frac{n_1}{n}\times 100=\frac{0.0625 mol}{0.1213mol}\times 100$

= 51.52%

Ammonia

= $\frac{n_2}{n}\times 100=\frac{0.0588 mol}{0.1213mol}\times 100$

= 48.47%