Final answer:
Using Gay-Lussac's Law and the formula P1/T1 = P2/T2, we find that a pressure increase in the can from 360 kPa to 750 kPa results in a temperature of 345.75°C (after correcting for the Kelvin to Celsius conversion).
Step-by-step explanation:
To determine the temperature of the same can if it contains gas at a pressure of 750 kPa, we will use Gay-Lussac's Law, which describes how the pressure of a gas tends to increase as the temperature increases, when the volume is held constant. The formula for Gay-Lussac's Law is P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
Given that the initial condition is 360 kPa at 24°C (add 273 to convert to Kelvin), we have P1 = 360 kPa and T1 = 297 K (since 24 + 273 = 297). The final pressure P2 is 750 kPa. We can now solve for T2.
Using the formula, we get: 360/297 = 750/T2, T2 = (750 * 297) / 360. After doing the math, we find that T2 = 618.75 K. To convert Kelvin back to Celsius, we subtract 273 from 618.75 K, resulting in T2 = 345.75°C.
Therefore, if the container now has a pressure of 750 kPa, the temperature inside the container would be 345.75°C, which is option (c) 67.2°C if the figure was meant to be in Celsius directly (which seems to be a typo since 67.2°C does not match our calculation).