Considering the Gay-Lussac's law, if the absolute pressure of a gas is increased from 3 atm to 4 atm at constant volume, then the absolute temperature of the gas will increase by [tex]\frac{4}{3}[/tex].
Gay-Lussac's law states that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.
In summary, when there is a constant volume, with increasing temperature, the pressure of the gas increases. And when the temperature is decreased, gas pressure decreases.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T} =k[/tex]
Studying an initial state 1 and a final state 2, it is fulfilled:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
In this case, you know:
Replacing in Gay-Lussac's law:
[tex]\frac{3 atm}{T1} =\frac{4 atm}{T2}[/tex]
Solving:
[tex]T2\frac{3 atm}{T1} =4 atm[/tex]
[tex]T2=\frac{4 atm}{\frac{3 atm}{T1} }[/tex]
[tex]T2=\frac{4 atm}{3 atm }T1[/tex]
[tex]T2=\frac{4 }{3 }T1[/tex]
Finally, if the absolute pressure of a gas is increased from 3 atm to 4 atm at constant volume, then the absolute temperature of the gas will increase by [tex]\frac{4}{3}[/tex].
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