What is gay lussac law? Explain with two examples.
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Answered by
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Gay-Lussac's gas law is a special case of the ideal gas law where the volume of the gas is held constant. When the volume is held constant, the pressure exerted by a gas is directly proportional to the absolute temperature of the gas. This example problem uses Gay-Lussac's law to find the pressure of a heated container.
GAY-LUSSAC'S LAW EXAMPLE
A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C?
Solution
The cylinder's volume remains unchanged while the gas is heated so Gay-Lussac's gas law applies. Gay-Lussac's gas law can be expressed as
Pi/Ti = Pf/Tf
where
Pi and Ti are the initial pressure and absolute temperatures
Pf and Tf are the final pressure and absolute temperature
First, convert the temperatures to absolute temperatures.
Ti = 27 °C = 27 + 273 K = 300 K
Tf = 77 °C = 77 + 273 K = 350 K
Use these values in Gay-Lussac's equation and solve for Pf.
Pf = PiTf/Ti
Pf = (6 atm)x(350K)/(300 K)
Pf = 7 atm
Answer:
The pressure will increase to 7 atm after heating the gas from 27 °C to 77 °C.
ANOTHER GAY-LUSSAC'S EXAMPLE
Find the temperature in Celsius needed to change the pressure of 10.0 liters of a gas that has a pressure of 97.0 kPa at 25°C to standard pressure. Standard pressure is 101.325 kPa.
First convert 25°C to Kelvin (298K).
Insert the numbers into the equation to get:
97.0 kPa / 298 K = 101.325 kPa / x
solving for x:
x = (101.325 kPa)(298 K)/(97.0 kPa)
x = 311.3 K
Subtract 273 to get the answer in Celsius.
x = 38.3°C
GAY-LUSSAC'S LAW EXAMPLE
A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C?
Solution
The cylinder's volume remains unchanged while the gas is heated so Gay-Lussac's gas law applies. Gay-Lussac's gas law can be expressed as
Pi/Ti = Pf/Tf
where
Pi and Ti are the initial pressure and absolute temperatures
Pf and Tf are the final pressure and absolute temperature
First, convert the temperatures to absolute temperatures.
Ti = 27 °C = 27 + 273 K = 300 K
Tf = 77 °C = 77 + 273 K = 350 K
Use these values in Gay-Lussac's equation and solve for Pf.
Pf = PiTf/Ti
Pf = (6 atm)x(350K)/(300 K)
Pf = 7 atm
Answer:
The pressure will increase to 7 atm after heating the gas from 27 °C to 77 °C.
ANOTHER GAY-LUSSAC'S EXAMPLE
Find the temperature in Celsius needed to change the pressure of 10.0 liters of a gas that has a pressure of 97.0 kPa at 25°C to standard pressure. Standard pressure is 101.325 kPa.
First convert 25°C to Kelvin (298K).
Insert the numbers into the equation to get:
97.0 kPa / 298 K = 101.325 kPa / x
solving for x:
x = (101.325 kPa)(298 K)/(97.0 kPa)
x = 311.3 K
Subtract 273 to get the answer in Celsius.
x = 38.3°C
Answered by
33
Answer:
Gay-Lussac's law states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant.
This formula can be expressed as follows:
(P1/T1) = (P2/T2) Where: P1 is the initial pressure. P2 is final Pressure. T1 is initial temp. T2 is final Temp....
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