General Law of the Gaseous State

The General Law of the Gaseous State the combination of the Three Gas Laws is considered: Boyle's Law, Gay-Lussac's Law and Charles's Law. Each one is in charge of relating two of the fundamental variables: Pressure, Volume and Temperature.

The General Law of the Gaseous State establishes the constant Relationship between Pressure, Volume and Temperature, in the form of the Equation:

PV / T = P’V ’/ T’

It means that the Pressure-Volume vs. Temperature Ratio will have the same value both at the beginning and at the end of a process involving gas. Such a process can be an expansion or a contraction.

Characteristics and Properties of Gases

Knowing that gases are made up of fast-moving molecules, we can understand why they act the way they do. If we descend into a deep mine or go up in an elevator, our eardrums respond to the change in altitude.

At high altitudes, the air molecules are farther apart, and at the depth of a mine they are closer together than at sea level. Assuming the temperatures are the same, the molecules move at the same speed, actually at the same speed. average speed, but in the mine they hit the eardrum in greater numbers than at sea level, in the same interval of weather.

This more intense bombardment (more pressure) of the eardrums is what produces in the ears the peculiar sensation of a descent into a deep mine.

Boyle's Law

Boyle's Law is one of the Gas Laws, and refers to the Variation in the Volume of a Gas Due to Pressure. Robert Boyle was the first to carefully study the effect of Pressure on the volumes of gases.

He observed that all gases behave in the same way when subjected to pressure changes, provided that the Temperature stay constant.

It can be stated as follows:

"The volume of all dry gas, at constant temperature, varies inversely to the pressure to which it is subjected"

It can be expressed mathematically as follows:

V varies as 1 / P

V = k (constant) * 1 / P

Or V * P = k

It is therefore also expressed:

"For any mass of dry gas at Constant Temperature, the product of Volume and Pressure is constant."

Charles Law

Charles studied the expansion of gases and showed that, holding pressure constant, all gases expand to an equal extent when heated by a specified number of degrees.

If a volume of gas is measured at 32 ° F and the temperature is raised to 33 ° F without varying the Pressure, the increase in Volume is equal to 1/492 of the original.

Charles's Law has as a mathematical expression:

V / T = V ’/ T’

It indicates that the relationship between Volume and Temperature is the same, both in an initial state and in the final state. This if the Constant pressure.

Gay-Lussac's Law

Gay-Lussac enunciated the Law that establishes how Pressure and Temperature are related when maintained constant the volume that the gas occupies.

When the pressure is low, the gas molecules will be more agitated. This is related to a high temperature. On the other hand, a higher pressure will compact the molecules and the system will cool down.

Gay Lussac's Law is expressed mathematically as:

P / T = P ’/ T’

General Law of the Gaseous State

Whenever a given mass of gas is measured, note not only the Volume, but also the pressure and temperature at which the measurement was made. It is often necessary to calculate the Volume at NTP (Normal Temperature and Pressure) conditions, when the volume is given under conditions other than these.

The General Law of the Gaseous State takes into account all the variables as fluctuating from one state of equilibrium to another, without one of them being constant.

PV / T = P’V ’/ T’

It continues to be established that the Relationship of these three variables is constant: Pressure-Volume between Temperature.

Examples of the General Law of the Gaseous State

1.-A quantity of gas occupies 300ml at 283K and 750mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 750mmHg

V = 300ml

T = 283K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (750mmHg) (300ml) (273K) / (760mmHg) (283K)

V ’= 286 ml

2.-A quantity of gas occupies 250ml at 343K and 740mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 740mmHg

V = 250ml

T = 343K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (740mmHg) (250ml) (273K) / (760mmHg) (343K)

V ’= 194 ml

3.-A quantity of gas occupies 100ml at 453K and 770mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 770mmHg

V = 100ml

T = 453K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (770mmHg) (100ml) (273K) / (760mmHg) (453K)

V ’= 61 ml

4.-A quantity of gas occupies 1500ml at 293K and 745mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 745mmHg

V = 1500ml

T = 293K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (745mmHg) (1500ml) (273K) / (760mmHg) (293K)

V ’= 1370 ml

5.-A quantity of gas occupies 2400ml at 323K and 767mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 767mmHg

V = 2400ml

T = 323K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (767mmHg) (2400ml) (273K) / (760mmHg) (323K)

V ’= 2047 ml

6.-A quantity of gas occupies 1250ml at 653K and 800mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 800mmHg

V = 1250ml

T = 653K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (800mmHg) (1250ml) (273K) / (760mmHg) (653K)

V ’= 550 ml

7.-A quantity of gas occupies 890ml at 393K and 810mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 810mmHg

V = 890ml

T = 393K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (810mmHg) (890ml) (273K) / (760mmHg) (393K)

V ’= 659 ml

8.-A quantity of gas occupies 320ml at 233K and 820 mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 820mmHg

V = 320ml

T = 233K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (820mmHg) (320ml) (273K) / (760mmHg) (233K)

V ’= 404 ml

9.-A quantity of gas occupies 1210ml at 413K and 795mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 795mmHg

V = 1210ml

T = 413K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (795mmHg) (1210ml) (273K) / (760mmHg) (413K)

V ’= 837 ml

10.-A quantity of gas occupies 900ml at 288K and 725mmHg of Pressure. Find the Volume in normal conditions: 273K and 760mmHg.

P = 725mmHg

V = 900ml

T = 288K

P ’= 760mmHg

V ’=?

T ’= 273K

PV / T = P’V ’/ T’

V ’= (P V T’) / (P ’T)

V ’= (725mmHg) (900ml) (273K) / (760mmHg) (288K)

V ’= 814 ml