## Section 5: Boyle's and Charles' Laws

Boyle's Law
Practice Problems
Charles's Law
Practice Problems

#### Boyle's Law

In the 17th century, an Irish chemist named Robert Boyle began to do quantitative experiments on gases. Using a gigantic J-shaped tube that he had mounted on the side of his house, Boyle studied the relationship between the pressure of the trapped gas and its volume. After collecting all of his data, Boyle noticed that the product of the pressure and volume for the gas was constant. This is represented by the equation PV = k, where P is pressure, V is volume, and k is a constant number at a specific temperature for a given sample of air. Boyle's law can be used to predict the new volume of a gas when the pressure is changed (at constant temperature).

Practice problem:

If you had a gas that exerted 10 atm of pressure and took up a space of 3 liters, and you decided to expand the space to 7 liters, what would be the new pressure? Assume that temperature remains constant.

Answer The equation PV = k will give you what the constant for this gas is. However, since you are using the same gas, the constant will be the same. Thus, the original product of pressure and volume will equal the new product of pressure and volume. This is represented by:

P1V1 = P2V2

where P1 and V1 are the pressure and volume before expanding (respectively)

and P2 and V2 are the pressure and volume after expanding.

So, P1 = 10 atm, V1 = 3 L, V2 = 7 L, and you need to solve for P2, the new pressure. Plug the numbers into the equation, and you have (10 atm) x (3 L) = (P2) x (7 L). P2 = 4.3 atm.

#### Charles's Law

In 1787, a French physicist named Jacques Charles found that the volume of a gas increases linearly with its temperature, at constant pressure. When graphed, using volume on the y-axis and temperature on the x-axis, a straight line develops. In all gases, when the line is extrapolated so volume = 0, the temperature is always -273.2 C. This point is defined as 0 K, on the Kelvin temperature scale, and is called absolute zero. In physics, you will learn that it is impossible to attain absolute zero, though temperatures as low as .0001 K have been attained in laboratories. The relationship between the Kelvin and Celsius temperature scales is as follows: K = 0 C + 273. In Charles's Law, it is important to remember that you must convert to Kelvins before doing the problem (teachers love to catch you on this on your tests). Charles's Law is represented by the equation V = bT, where V = volume, b = a proportionality constant, and T = temperature (IN KELVINS!!).

Practice problem:

If you took a balloon outside that was originally at 20 C at 2 L in volume, and it heated up to 29 C, what would its volume be?

Examples The equation V = bT can be rearranged to V/T = b. Now, you know that since you are dealing with the same gas at constant pressure (always assume constant pressure, unless if the problem tells you otherwise), the constant, b, will remain the same. So:

V1 / T1 = b = V2 / T2 or V1 / T1 = V2 / T2.

V1 = 2 L, T1 = 20 C, T2 = 29 C, and you must solve for V2. Wait!! You have to convert the temperatures to Kelvin first!! So:

T1 = 20 C = 20 C + 273 = 293 K
and T2 = 29 C = 29 C + 273 = 302 K

Now, you can plug in the numbers and solve for V2. (2 L / 293 K) = (V2 / 302 K), and V2 = 2.1 L