Subtitles section Play video Print subtitles For a gas, temperature and volume are directly proportional. Keeping everything else constant, as the temperature of a gas goes up, its volume goes up. As the temperature of a gas goes down, its volume goes down. If you heat up a gas, it expands - the gas particles move faster, and they take up more space. Imagine a balloon, that expands when gas particles bang against the sides. (by expanding, that keeps the pressure constant). The faster the gas particles move, the more they will push on the sides of the balloon, expanding it. If, on the other hand, you cool the gas down - you put your balloon on ice - that slows the particles of gas down, so the balloon will contract. Jacques Charles is credited with this Gas Law relating temperature and volume, although he didn’t publish it - he may have been too busy taking rides in hot air balloons. A colleague, Joseph Louis Gay-Lussac, published it and very honorably gave Charles credit. Charles’s Law says that for a given amount of gas, at fixed pressure, volume and temperature are directly proportional. V ∝ T You can write this mathematically as V = kT where V = volume, T = temperature in Kelvin, and k = is a proportionality constant. We can rearrange this equation so it reads V/T = k, or the ratio of volume to temperature is a constant, k. Very often, Charles’s law is used to compare two situations, a “before” and an “after.” In that case, you can say V1 /T1= k, and V2/T2 = k, so you can write Charles’s law as V1 /T1= V2/T2. Let’s see an example. A hot air balloon has a volume of 2,800 m3 at 99 C°. What is the volume if the air cools to 80 C°? We’ll write Charles’s Law in the “before and after” form: V1 /T1= V2/T2. We substitute in what we know - remember to convert temperatures to Kelvin: Kelvin = C° + 273.15. T1= 372.15 Kelvin, T2 = 353.15 Kelvin 2,800 m3/ 372.15 K = V2/ 353.15 K solve for V2 (multiply both sides by 353.15K) (353.15 K) (2,800 m3)/ 372.15 K = V2 V2 = 2657 m3 Here’s another example: At 0 C°, a gas occupies 22.4L. How hot must the gas be, in Celsius, to reach a volume of 25.0 L? V1 /T1= V2/T2. Convert temperature to Kelvin: Kelvin = C° + 273.15. T1 = 273.15 K Substituting in what we know: 22.4L/273.15K = 25.0L/T2 Solve for T2 (I like to “cross multiply”) (22.4L) T2 = (273.15 K)(25.0L) T2 = (273.15K) (25.0L)/22.4L T2 = 304.9 K Convert to C°: C° = Kelvin - 273.15 T2 = 31.7 C° Charles’s law relates temperature and volume for a gas, but there are other gas laws which relate the other essential variables associated with a gas. Boyle’s Law is the relationship between pressure and volume. Gay-Lussac’s Law is the relationship between pressure and temperature. And the combined gas law puts all 3 together: Temperature, Pressure, and Volume. Notice that to use any of these laws, the amount of gas must be constant. Avogadro’s Law describes the relationship between volume and the amount of a gas (usually in terms of n, the number of moles). When we combine all 4 laws, we get the Ideal Gas Law. To decide which of these gas laws to use when solving a problem, make a list of what information you have, and what information you need. If a variable doesn’t come up, or is held constant in the problem, you don’t need it in your equation.
B1 gas t2 volume kelvin temperature t1 Chemistry: Charles's Law (Gas Laws) with 2 examples | Homework Tutor 1 1 林宜悉 posted on 2020/03/06 More Share Save Report Video vocabulary