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Guy-Lussac's Gas Law Example

Ideal Gas Law Example Problem

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Guy-Lussac's gas law is a special case of the ideal gas law.

Guy-Lussac's gas law is a special case of the ideal gas law

Paul Taylor, Getty Images
Guy-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 Guy-Lussac's law to find the pressure of a heated container.

Problem:

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.

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