P1/T1 = P2/T2
125⁰C = 398.15 k
182⁰C = 455.15 k
1.22/398.15 = p2/455.15
p2= 1.39atm
the pressure of the gas be after the temperature change is 1.39 atm
Answer:
1.4 atm
Explanation:
According to Gay Lussac’s law, the pressure of gas is inversely proportional t its temperature, when the volume is constant
Initial temperature = 125 + 273 = 398 K
Final temperature = 182 + 273 = 455 K
Initial pressure = 1.22 atm
[tex]\frac{P1}{T1} = \frac{P2}{T2}[/tex]
1.22 atm / 398 K = P₂/ 455 K
P₂ = [tex]\frac{1.22 atm x 455 K}{398 K}[/tex] = 1.4 atm
