Answer:
T₂ = 182 K
Explanation:
Given that,
Initial pressure, P₁ = 120 kPa
Initial temperature, T₁ = 0˚C = 273 K
We need to find the final temperature when the pressure is 80 kPa.
We know that, Gay Lussac's Formula is :
[tex]\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\\\\T_2=\dfrac{P_2T_1}{P_1}\\\\T_2=\dfrac{80\ kPa\times 273}{120\ kPa}\\\\T_2=182\ K[/tex]
So, the new temperature is equal to 182 K.