Explanation:
The given data is as follows.
[tex]V_{1}[/tex] = 250 mL, [tex]V_{2}[/tex] = 750 mL
[tex]T_{1}[/tex] = [tex]35^{o}C[/tex] = 35 + 273 K = 308 K
[tex]T_{2}[/tex] = 35 + 273 K = 308 K
[tex]P_{1}[/tex] = 0.55 atm, [tex]P_{2}[/tex] = 1.5 atm
P = ? , V = 10.0 L
Since, temperature is constant.
So, [tex]P_{1}V_{1} + P_{2}V_{2}[/tex] = PV
Now, putting the given values into the above formula as follows.
[tex]P_{1}V_{1} + P_{2}V_{2}[/tex] = PV
[tex]0.55 atm \times 250 mL + 1.5 atm \times 1.5 atm[/tex] = [tex]P \times 10.0 L[/tex]
P = 0.126 atm
As, 1 atm = 760 torr. So, [tex]0.126 atm \times \frac{760 torr}{1 atm}[/tex] = 95.76 torr.
Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.
