The ideal gas law gives the relationships among the four variables of temperature, pressure, volume and molar amount for gaseous substances at a given set of conditions. **"The ideal gas known also as the equation of state for ideal gases.****"**

An ideal gas is a gas in which every molecule behaves independently of every other molecule and has no excluded volume.**Ideal Gas Law Formula** is mathematically expressed as

An ideal gas is a gas in which every molecule behaves independently of every other molecule and has no excluded volume.

The ideal gas contains four variables and the constant R. Given any three, the fourth variable can be found. In addition this law can be used to find the molar mass or molecular weight if the mass of a gas is given.

An ideal gas is a hypothetical gas dreamed by chemists and students because it would be much easier if things like inter molecular forces do not exist to complicate the simple Ideal Gas Law. Ideal gases are essentially point masses moving in constant, random, straight-line motion. Its behavior is described by the assumptions listed in the Kinetic-Molecular Theory of Gases.

An ideal gas is a hypothetical gas dreamed by chemists and students because it would be much easier if things like inter molecular forces do not exist to complicate the simple Ideal Gas Law. Ideal gases are essentially point masses moving in constant, random, straight-line motion. Its behavior is described by the assumptions listed in the Kinetic-Molecular Theory of Gases.

The ideal gas is valid for a single individual gas. If a mixture of gases is present the ideal gas law is valid for each individual gas.

Related Calculators | |

Ideal Gas Law Calculator | Combined Gas Law Calculator |

Calculating Molar Mass of a Gas | gas pressure drop calculator |

Solved problems based on ideal gas law are given below. ### Solved Examples

**Question 1: **One mole of CH_{4} gas occupies 20.0L at 1.00atm pressure. What is the temperature of the gas in kelvin?

** Solution: **

Solve the ideal gas law for T and plug in the given values.

T = $\frac{PV}{nR}$

T = $\frac{PV}{n}$ $\times$ $\frac{1}{R}$

T = $\frac{[1.00atm][20.0L]}{1.00mol}$ $\times$ $\frac{mol.K}{0.0821L.atm}$

T = 244K

Note that we calculated the temperature for 1.00 mol of CH_{4} gas under these conditions. The answer would be the same for 1.00 mol of CO_{2}, N_{2}, NH_{3} or any other gas under these conditions.

**Question 2: **If there is 5.0g of CO_{2} gas in a 10 L cylinder at 25^{o}C, what is the gas pressure within the cylinder?

** Solution: **

We are given the quantity of CO_{2} in grams but to use the ideal gas law, we must express the quantity in moles. Therefore we must first convert grams of CO_{2} to moles, and then use this value in the ideal gas law.

To convert from grams to moles we use the conversion factor 1 mol CO_{2} = 44g.

5.0g CO_{2} $\times$ $\frac{1\ mol\ CO_{2}}{44g\ CO_{2}}$ = 0.11 mol CO_{2}

We now use this value in the ideal gas equation to solve for the pressure of the gas.

P = $\frac{nRT}{V}$

P = $\frac{[0.11\ mol\ CO_{2}][298K]}{10L}$ $\times$ $\frac{0.0821L.atm}{mol.K}$

P = 0.27atm

Solve the ideal gas law for T and plug in the given values.

T = $\frac{PV}{nR}$

T = $\frac{PV}{n}$ $\times$ $\frac{1}{R}$

T = $\frac{[1.00atm][20.0L]}{1.00mol}$ $\times$ $\frac{mol.K}{0.0821L.atm}$

T = 244K

Note that we calculated the temperature for 1.00 mol of CH

We are given the quantity of CO

To convert from grams to moles we use the conversion factor 1 mol CO

5.0g CO

We now use this value in the ideal gas equation to solve for the pressure of the gas.

P = $\frac{nRT}{V}$

P = $\frac{[0.11\ mol\ CO_{2}][298K]}{10L}$ $\times$ $\frac{0.0821L.atm}{mol.K}$

P = 0.27atm

More topics in Ideal Gas Law Formula | |

Vapor Pressure Formula | Combined Gas Law Formula |

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