Answer: The pressure at 273 K is 130 kPa.
Solution:- From Gay-Lussac's law, "At constant volume, the pressure of a given mass of a gas is directly proportional to the kelvin temperature."
The equation used for this is written as:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
Where, [tex]P_1[/tex] is the pressure at [tex]T_1[/tex] temperature and [tex]P_2[/tex] is the pressure at [tex]T_2[/tex] temperature.
Temperatures are already given in kelvin so no need to do the conversion. Let's plug in the values and calculate the new pressure.
[tex]\frac{340kPa}{713K}=\frac{O_2}{273K}[/tex]
On rearranging this for new pressure:
[tex]P_2=\frac{340kPa*273K}{713K}[/tex]
[tex]P_2=130kPa[/tex]
