Question

15. An apparatus consists of a 4.0 dm3

Answer 1

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$x_{1}$ be the partial pressure of the nitrogen gas

(P$x_{1}$)×14=3212

∴P$x_{1}$=219.429kPa

∴Partial pressure of nitrogen gas is 219.429kPa

Initially for argon gas PV=47.2×10=472

let P$x_{2}$ be the partial pressure of the argon gas

(P$x_{2}$)×14=472

∴P$x_{2}$=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$x_{1}$+P$x_{2}$

                          =219.429+33.719

Total pressure    =253.143kPa

Answer 2

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  

$\large{\boxed{\bold{PV=nRT}}}$

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

$\rm n=\dfrac{7.925\times 4}{0.08205\times 298}\\\\n=1.289$

2. mol Ar

P = 47.2 kPa = 0.466 atm

V = 10 L

$\rm n=\dfrac{0.466\times 10}{0.08205\times 298}\\\\n=0.191$

Then :

V total = 14 L

• Find partial pressure of each gas after valve is opened

1. P N₂

$\rm P_{N_2}=\dfrac{1.289\times 0.08205\times 298}{14}\\\\P_{N_2}=2.251\:atm$

2. P Ar

$\rm P_{Ar}=\dfrac{0.191\times 0.08205\times 298}{14}\\\\P_{Ar}=0.3336\:atm$

P total = P N₂ + P Ar

P total = 2.251 + 0.3336

P total = 2.585 atm

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