Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
The new pressure of the gas is 2.25 atm.
We know that the pressure of a given mass of gas is directly proportional to the number of moles of the gas.
Mathematically;
P ∝ n
For two masses of an ideal gas;
P1/n1 = P2/n2
Where
P1 = initial pressure
P2 = Final pressure
n1 = number of moles of gas initially present
n2 = Final moles of gas present
Let the number of moles of the Ar gas be n
Given that;
P1 = 0.75atm
n1 = n
P2 = ?
n2 = 2n
Substituting into the given equation and making P2 the subject of the formula;
P2 = P1n2/n1
P2 = 0.75atm ×2n/n
P2 = 1.5 atm
Since the original mixture was equimolar, the new total pressure of the gas is partial pressure of N2 + new partial pressure of Ar = 0.75 atm + 1.5 atm = 2.25 atm
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