Final answer:
To increase the pressure of a gas cylinder at room temperature to 1.2P₁, using Gay-Lussac's Law, the temperature must be raised to approximately 78.63°C.
Step-by-step explanation:
To determine the new temperature needed to increase the pressure of a gas to 1.2P₁, we can use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume and amount of gas are kept constant.
This mathematical relationship can be represented by the equation \( \frac{P1}{T1} = \frac{P2}{T2} \), where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.
Given that the initial temperature is room temperature (20°C) and the desired pressure is 1.2 times the initial pressure P1, we first convert the Celsius temperature to Kelvin by adding 273.15. So the initial temperature T1 in Kelvin is 293.15 K. By rearranging Gay-Lussac's Law to solve for T2, we get \( T2 = T1 \times \frac{P2}{P1} \).
Substituting the known values, \( T2 = 293.15 K \times \frac{1.2P1}{P1} \), we can cancel out P1, which gives us \( T2 = 293.15 K \times 1.2 \). Finally, by multiplying, we find that T2 equals 351.78 K.
To get the answer back in Celsius, we subtract 273.15 from the Kelvin temperature, resulting in 78.63°C. Hence, the gas temperature must be increased from 20°C to approximately 78.63°C for the pressure to become 1.2P₁.