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:
\(\frac{P_1}{T_1}=\frac{P_2}{T_2}\)
Where, \(P_1\) is the pressure at \(T_1\) temperature and \(P_2\) is the pressure at \(T_2\) 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.
\(\frac{340kPa}{713K}=\frac{O_2}{273K}\)
On rearranging this for new pressure:
\(P_2=\frac{340kPa*273K}{713K}\)
\(P_2=130kPa\)
