I have 130 L of gas in a piston at a temperature of 250°C. If I cool the gas until the volume decreases to 85 L, what will the temperature of the gas be?

This is a Charles' Law problem: V1/T1 = V2/T2. As the temperature of a fixed mass of gas decreases at a constant pressure, the volume of the gas should also decrease proportionally. Here, we are given the new volume of the gas after cooling, and we want to determine to what temperature the gas was cooled.

To use Charles' Law, the temperature must be in Kelvin (x °C = x + 273.15 K). We want to solve Charles' Law for T2, which we can obtain by rearranging the equation into T2 = V2T1/V1. Given V1 = 130 L, T1 = 250 °C (523.15 K), and V2 = 85 L:

T2 = (85 L)(523.15 K)/(130 L) = 342 K or 68.9 °C. If sig figs are to be considered, since all the values in the question are given to two sig figs, the answer to two sig figs would be either 340 K or 69 °C.  

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

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i believe it is 40–40–20%

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