Gay-Lussac’s+Law

Gay-Lussac’s Law: P1/T1 = P2/T2 Gay-Lussac’s Law states the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in Kelvin. The relationship is similar to the volume-temperature relationship of Charles’s Law where as the temperature is increased, pressure increases as well. Problem: The gas in a container is at a pressure of 3.0 atm at 25 C. Directions on the container warn the user not to keep in a place where the temperature exceeds 52 C. What would the gas pressure in the container be at 52 C? After reading, you should be able to tell that Gay-Lussac’s Law should be used to solve this problem because it is the only equation which directly relates pressure and temperature. P1/T1 = P2/T2 P1 = initial pressure T1 = initial temperature P2 = final pressure T2 = final temperature

Start by identifying your givens and determining what you are solving for. The problem gives you both P1 and T1, 3.0 atm and 25 C and also T2 which is 52 C. The problem is asking you to solve for P2, the pressure within the container at 52 C. Before solving the equation, you need to make sure that the temperature in degrees Celsius is converted to Kelvin. To complete this conversion, add 273 to the temperature in C. 25 C + 273 = 298 K (T1) 52 C + 273 = 325 K (T2) Now simply put the values into the equation and solve the problem. 3.0 atm/298 K = P2/325 K To isolate and solve for P2, multiply both sides by 325 K so that it will cancel and leave P2 by itself on the right side of the equation. 3.0 atm/298 K * (325 K) = P2/325 K * (325 K)  P2 = 3.0 atm/298 K * (325 K)  P2 = 3.27 atm The problem is solved! :]]

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