When the pressure that a gas exerts on a sealed container changes from 811 mm Hg to 415 mm Hg, the temperature changes from 33.0°C to _ °C?
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Answer:
The final temperature of the gas will be 116.5 °C.
Explanation:
It can be calculated by Gay-Lussac. According to this law, the pressure is directly proportional to the temperature of the gas.
Initial pressure = P1 = 811 mmHg = 1.06 atm
Initial temperature = T1 = 33°C = [33 + 273.15]K = 306.15K
Final pressure = 415mmHg = P2 = 0.55 atm
T2= ?
P1 / T1 = P2 /T2
1.06 / 306.15 = 0.55 / T2
T2 = 157K
273.15 - 157= 116.5 degree Celsius
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