Gay Lussacs Law Formulas - Equation and Problem Solved with Example
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The law states that at a constant volume, the pressure (in atm) of a given amount of gas is directly proportional to the temperature (in Kelvin).
Mathematically,
P \[ \propto \] T at constant volume,V P/T = k where k is a constant
Which implies, P_{1 }/ T_{1} = P_{2 }/ T_{2} Or graphically, When pressure of the gas is plotted against temperature at constant volume, we would obtain a straight line passing through the origin. At any point on the graph the ratio of P : T will always be constant.
Example: A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased to 50.0 °C? Solution: From Gay Lussac’s Law we know that at constant volume (30 L in this question), P_{1 }/ T_{1} = P_{2 }/ T_{2} Convert given temperature in degree Celsius to Kelvin T_{1} = 20 + 273 T_{1} = 293 K
T_{2} = 50 + 273 T_{2} = 323 K Let, P_{1} = 3.00 atm and P_{2} = x; Then, 3 / 293 = x / 323 Which implies, x = 3.31 atm
Question: A gas has a pressure of 699.0 mmHg at 40.0 °C. What is the temperature at standard pressure Options: (a) 310 K (b) 380 K (c) 400 K (d) 340 K Answer: (d) Solution: P_{1 }/ T_{1} = P_{2 }/ T_{2} 699.0 / 313 = 760 / x x = 340 K (notice how pressure is not converted to atm, it is because the conversion factor gets cancelled from both sides of the equation)