Answer :
Answer:
627.4 mmHg
Explanation:
From Gay-Lussac law:
[tex] \frac{p1}{t1} = \frac{p2}{t2} [/tex]
p1 = 760 mmHg
t1 = 343 K
t2=283.15 K
[tex]p2 = p1 \times \frac{t2}{t1} [/tex]
[tex]p2 = 760 \times \frac{283.15}{343.0} [/tex]
[tex]p2 = 627.4 \: mmhg[/tex]
If a gas is cooled from 343.0 K to 283.14 K and the volume is kept constant the final pressure would be 627.4 mm Hg if the original pressure was 760.0 mm Hg.
Gay-Lussac's Law
Gay-Lussac's law states that the pressure of a given mass of gas at a constant volume varies directly with the temperature of the gas. Gay-Lussac's law defined the relationship between pressure and temperature of a gas when kept at a constant volume.
According to the Gay-Lussac's Law
P ∝ T (When V = constant)
or, P = kT
or, [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
Here,
[tex]P_{1}[/tex] = 760.0 mm Hg,
[tex]T_{1}[/tex]= 343.0 K
[tex]T_{2}[/tex] = 283.15 K
Now put the values in above formula, we get
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex]
[tex]\frac{760.0\ \text{mm Hg}}{343.0K} = \frac{P_2}{283.15\ K}[/tex]
[tex]P_{2} = \frac{760.0\ \text{mm Hg} \times 283.15\ K}{343.0\ K}[/tex]
[tex]P_{2} = \frac{215194\ \text{mm Hg}}{343}[/tex]
[tex]P_{2}[/tex] = 627.4 mm Hg
Thus, we can say that if a gas is cooled from 343.0 K to 283.14 K and the volume is kept constant the final pressure would be 627.4 mm Hg if the original pressure was 760.0 mm Hg.
Learn more about Gay-Lussac's Law here: https://brainly.com/question/24691513
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