Question

2. The temperature of a gas is increased from 125⁰C to 182⁰C inside of a rigid container. The original pressure of the gas was 1.22atm, what will the pressure of the gas be after the temperature change?

Answer 1

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 2

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

$\frac{P1}{T1} = \frac{P2}{T2}$

1.22 atm / 398 K = P₂/ 455 K

P₂ = $\frac{1.22 atm x 455 K}{398 K}$ = 1.4 atm