What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of CO2 contained in a 9 dm cube flask at 27 degree C?
Answers
Answered by
303
Hi Sister,
Here is your answer,
Moles of CH₄,
nCH₄= Mass of CH₄/Molar mass of CH₄ [ Molar mass of CH₄] = 12+4×1= 16
3.2/16 = 0.2 mol
Moles of CO₂
nCO₂ = 4.4/44 = 0.1 mol [ Molar mass of CO₂ = 12+2 ×16=44]
Total moles = 0.2 + 0.1 = 0.3 mol
pV=nRT
Pressure, P=nRT/V
0.3 mol × 0.0821 dm³ atm K power -1 mol power -1 × 300 K/ 9 dm³ = 0.0821 atm
In terms of SI unit
Pressure,p = 0.3 mol × 8.314 Pa m³ K power -1 mol power -1 × 300/ 9×10⁻³ m³
Therefore, p= 8.314×10 power 4 Pa.
Hope it helps you!
Here is your answer,
Moles of CH₄,
nCH₄= Mass of CH₄/Molar mass of CH₄ [ Molar mass of CH₄] = 12+4×1= 16
3.2/16 = 0.2 mol
Moles of CO₂
nCO₂ = 4.4/44 = 0.1 mol [ Molar mass of CO₂ = 12+2 ×16=44]
Total moles = 0.2 + 0.1 = 0.3 mol
pV=nRT
Pressure, P=nRT/V
0.3 mol × 0.0821 dm³ atm K power -1 mol power -1 × 300 K/ 9 dm³ = 0.0821 atm
In terms of SI unit
Pressure,p = 0.3 mol × 8.314 Pa m³ K power -1 mol power -1 × 300/ 9×10⁻³ m³
Therefore, p= 8.314×10 power 4 Pa.
Hope it helps you!
kvnmurty:
is the volume right? Chk answer
need to modify
Sir i have checked it throughly i am the getting the same answer !
dm = 0.9 m... dm^3 = 0.9^3 m^3
sir you mean to say my answer is wrong sir?
Think so
actually i have asked my chemistry teacher she told that volume is correct !
you are right @cbsedelhi
volume should be taken in dm³ not in m³ coz 1dm³ = 1 l (standard unit )
ask sir to correct his answer"
Answered by
193
Let's apply Ideal gas law: P V = n R T
T = 300°K.
V = 9 dm^3 = 9 * 10⁻³ m³ or 9 L
Molecular mass of Methane: CH4 =16.
of Carbon Dioxide CO2 = 44.
Total Number of moles of the gases in the mixture:
n = n1 + n2
= 3.2/16 + 4.4/44
= 0.3 mol
R = 8.318 J/°K
P = n R T / V
= 0.3 mol × 8.314 J/°K/mol × 300°K / 0.009m³
P = 0.8314 bar or 8.314 * 10⁴ Pa or 0.83 atm
We can also calculate the partial pressure exerted by Methane and Carbon DIoxide individually. .
P1 = n1/(n1+n2) * P
P2 = n2/(n1+n2) *P
T = 300°K.
V = 9 dm^3 = 9 * 10⁻³ m³ or 9 L
Molecular mass of Methane: CH4 =16.
of Carbon Dioxide CO2 = 44.
Total Number of moles of the gases in the mixture:
n = n1 + n2
= 3.2/16 + 4.4/44
= 0.3 mol
R = 8.318 J/°K
P = n R T / V
= 0.3 mol × 8.314 J/°K/mol × 300°K / 0.009m³
P = 0.8314 bar or 8.314 * 10⁴ Pa or 0.83 atm
We can also calculate the partial pressure exerted by Methane and Carbon DIoxide individually. .
P1 = n1/(n1+n2) * P
P2 = n2/(n1+n2) *P
:-)
^_^
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