Answer:
[tex]P_N=1.33atm\\\\P_O=0.67atm[/tex]
[tex]P_T=2.00 atm[/tex]
Explanation:
Hello.
In this case, the first step is to compute the moles of nitrogen and oxygen given their initial P, T and V conditions via the ideal gas equation:
[tex]n_{N}=\frac{P_{N}V_{N}}{RT} =\frac{2.00atm*24.0L}{0.082\frac{atm*L}{mol*K}*273K}\\ \\n_{N}=2.144molN\\\\n_{O}=\frac{P_{O}V_{O}}{RT} =\frac{2.00atm*12.0L}{0.082\frac{atm*L}{mol*K}*273K}\\ \\n_{O}=1.072molO[/tex]
After that, since the total volume now, once the mixture is formed is the addition between the initial volumes (12.0 L + 24.0 L) is 36.0 L, the partial pressure of each gas turns out:
[tex]P_N=\frac{2.144mol*0.082\frac{atm*L}{mol*K}*273K}{36.0L}\\\\P_N=1.33atm\\\\P_O=\frac{1.072mol*0.082\frac{atm*L}{mol*K}*273K}{36.0L}\\\\P_O=0.67atm[/tex]
Thus, the final total pressure is:
[tex]P_T= P_N+P_O=1.33atm+0.67atm\\\\P_T=2.00 atm[/tex]
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