At 400 K, a volume of gas has a pressure of 0.40 atmospheres. What is the pressure of this gas at 273 K?
Answer (2)
The pressure of the sample of gas at 273 K is equal to 0.273 atm assuming the volume is constant.
What is Gay Lussac's law?
Gay-Lussac's law can be described as that when the volume of the gas remains constant then the pressure (P) of the gas and the** absolute temperature **(T) in a direct relationship.
**Gay Lussca's law **can be described as mentioned below:
P/T = k
The pressure of gas is directly proportional to absolute temperature of the gas.
P ∝ T ( Volume of gas is constant)
or, P₁/T₁ = P₂/T₂ .................(1)
Given, the** initial temperature** of the gas, T₁ = 400 K
The** final temperature **of the gas, T₂ = 273 K
The initial pressure of the gas, P₁ = 0.40 atm
The final pressure of the given gas can be calculated from **Gay Lussac's law **as:
P 2 = T 1 P 1 × T 2
P₂ = (0.40/400) ×273
P₂ = 0.273 atm
Therefore, the pressure of this gas at 273 K is 0.273 atm.
Learn more about** Gay Lussac's law**, here:
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Using Gay-Lussac's Law, the pressure of the gas can be calculated as 0.273 atm at 273 K. This relationship shows that pressure is directly proportional to temperature when volume is constant. Therefore, as temperature decreases, pressure also decreases.
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