Answer:
Explanation:
From the information given:
Step 1:
Determine the partial pressure of each gas at total Volume (V) = 4.0 L
So, using:
[tex]\text{The new partial pressure for }N_2 \ gas}[/tex]
[tex]P_1V_1=P_2V_2[/tex]
[tex]P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar[/tex]
[tex]\text{The new partial pressure for }Ar \ gas}[/tex]
[tex]P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar[/tex]
[tex]Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar[/tex]
Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.
According to Ideal gas Law.
[tex]2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm[/tex]
For moles Nā:
[tex]PV = nRT \\ \\ n = \dfrac{PV}{RT}[/tex]
[tex]n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}[/tex]
[tex]n = 0.08297 \ mol \ N_2[/tex]
For moles of Ar:
[tex]PV = nRT \\ \\ n = \dfrac{PV}{RT}[/tex]
[tex]n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}[/tex]
[tex]n = 0.2172 \ mol \ Ar[/tex]
[tex]\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}[/tex]
[tex]=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases[/tex]
Finally;
The final pressure of the mixture is:
[tex]PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}[/tex]
P = 2.217 atm
P ā 2.24 bar
