Respuesta :
Answer:
i· Partial pressure of nitrogen gas is 219.429kPa and partial pressure of argon gas is 33.714kPa ·
ii· Total pressure of the gas mixture is 253.143kPa·
Explanation:
solution for i :
Assuming that the given gases to be ideal,
so by ideal gas equation
PV=nRT
where,
P is the pressure of the gas
V is the volume occupied by the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas
Actually partial pressure of each gas is the pressure of the gas exerted when it occupies complete volume
As the temperature of both gases are same, the mixing process is an isothermal process
PV=constant
Initially for nitrogen gas PV=803×4=3212
let P[tex]x_{1}[/tex] be the partial pressure of the nitrogen gas
(P[tex]x_{1}[/tex])×14=3212
∴P[tex]x_{1}[/tex]=219.429kPa
∴Partial pressure of nitrogen gas is 219.429kPa
Initially for argon gas PV=47.2×10=472
let P[tex]x_{2}[/tex] be the partial pressure of the argon gas
(P[tex]x_{2}[/tex])×14=472
∴P[tex]x_{2}[/tex]=33.714kPa
∴Partial pressure of argon gas is 33.714kPa
solution for ii :
Total pressure of the gas mixture will be the sum of the partial pressures of each gas as
Partial pressure of the gas=(total pressure of the mixture)×(mole fraction of the gas)
∴Total pressure=P[tex]x_{1}[/tex]+P[tex]x_{2}[/tex]
=219.429+33.719
Total pressure =253.143kPa
Partial pressure : P N₂ = 2.251 atm; P Ar = 0.3336 atm
Total Pressure = 2.585 atm
Further explanation
Some of the laws regarding gas, can apply to ideal gas (volume expansion does not occur when the gas is heated), among others
- Boyle's law at constant T, P = 1 / V
- Charles's law, at constant P, V = T
- Avogadro's law, at constant P and T, V = n
So that the three laws can be combined into a single gas equation, the ideal gas equation
In general, the gas equation can be written
[tex]\large{\boxed{\bold{PV=nRT}}}[/tex]
where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08205 l.atm / mol K
T = temperature, Kelvin
- Find mol of each gas before valve is opened
1. mol N₂
P = 803 kPa = 7.925 atm
V = 4 L
T = 25 C = 298 K
[tex]\rm n=\dfrac{7.925\times 4}{0.08205\times 298}\\\\n=1.289[/tex]
2. mol Ar
P = 47.2 kPa = 0.466 atm
V = 10 L
[tex]\rm n=\dfrac{0.466\times 10}{0.08205\times 298}\\\\n=0.191[/tex]
Then :
V total = 14 L
- Find partial pressure of each gas after valve is opened
1. P N₂
[tex]\rm P_{N_2}=\dfrac{1.289\times 0.08205\times 298}{14}\\\\P_{N_2}=2.251\:atm[/tex]
2. P Ar
[tex]\rm P_{Ar}=\dfrac{0.191\times 0.08205\times 298}{14}\\\\P_{Ar}=0.3336\:atm[/tex]
P total = P N₂ + P Ar
P total = 2.251 + 0.3336
P total = 2.585 atm
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