# The Combined Cas Law

2021-05-12 02:48:20
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The combined gas law is a gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law.

This law states: The ratio between the pressure-volume product divided by the temperature of a system remains constant.

This can be stated mathematically as:

Where: p is the pressure, V is the volume, T is the temperature measured in kelvins, and k is a constant (with units of energy divided by temperature).

Reminder: 1atm= 760 torr = 101.3 kPa & Celsius to Kelvin= add 273 and Kelvin to Celsius= subtract 273

If the problem does not state which unit to give the result in, then make sure that temperature is converted into Kelvin and for the Pressure and Volume just make sure you stay constant and use the same unit on both sides of the equation.

Combination of 3 Laws:

Boyle's Law states that the pressure-volume product is constant:

In other words as external pressure on a gas increases the volume decreases, and vice versa.

Charles's Law shows that the volume is proportional to absolute temperature:

In other words as temperature increases the volume increases, and vice versa.

Gay-Lussac's Law says that the pressure is proportional to the absolute temperature:

In other words as temperature increases the pressure increases, and vice versa.

Where P is the pressure, V the volume and T the absolute temperature and of an ideal gas.

By combining (1) and either of (2) or (3) we can gain a new equation with P, V and T. Equation (2) is used in this example, and the subscript on the constant is dropped so that k = k2. We get the Combined Gas Law! (No specific scientist invented this law; rather three laws were just combined mathematically to come up with this law)

The combined gas law can be used to explain the mechanics where pressure, temperature, and volume are affected. For example: air conditioners, refrigerators and the formation of clouds.

Calculations with the Combined Gas Law:

For example, the pressure and temperature of a gas are changed to STP (101.3 kPa/00C) from 22.00C and 30.8 kPa. What will be the new volume if the original volume was 205 mL?

Solution:

First, we must convert degrees Celsius to Kelvins. This means that our original temperature was 295 K and our target temperature is 273 K.

Plug values into the formula

Given:

Volume (V) = 205 mL, Temperature (T) = 295 K, Pressure (P) = 30.8 kPa

Temperature1 (T1) = 273 K, Pressure1 (P1) = 101.3 kPa, the unknown is V1. Our equation is:

30.8 kPa x 205 mL / 295 K = 101.3 kPa x V1 / 273 K

Manipulating the equation, we get:

V1 = 30.8 kPa x 205 mL / 295 K x 273 K / 101.3 kPa

V1 = 57.68 mL, However, because our problem only was to three digits of precision, the answer is:

V1 = 57.7 mL: Therefore the new volume of the gas is 57.7 mL.

A couple of examples are the internal combustion engine. For example, a four stroke engine like your car operates on the principle of taking a volume of gas/air mixture, compressing it, igniting it, and pushing the exhaust out. The movement of the pistons moves the drive shaft..... Also, weather balloons are launched daily from weather stations across the country. The balloon begins at the earth at a certain P, T, and V and upon its accent all three of these variables change in response to the surroundings. Have the same topic and dont`t know what to write?
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