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# What is an Ideal Gas?

## Important Formula

What do the formula symbols mean?

## Pressure

Unit
This pressure is present in a confined system containing an ideal gas. According to the ideal gas law, the pressure increases when the temperature $$T$$ of the gas increases or the volume $$V$$ in which the gas is confined decreases.

## Volume

Unit
The volume of a closed system containing an ideal gas.

## Temperature

Unit
It is the absolute temperature (in Kelvin) of the gas in a closed system.

## Amount of substance

Unit
The amount of substance indirectly indicates the number of gas particles. It is related to the particle number $$N$$ by the Avogardo constant $$N_{\text A}$$: $$n = \frac{N}{N_{\text A}}$$.

## Gas constant

Unit
Molar gas constant (also called universal gas constant) is a physical constant from thermodynamics and has the following exact value: $$R ~=~ 8.314 \, 462 \, 618 \, 153 \, 24 \, \frac{\mathrm J}{\mathrm{mol} \, \mathrm{K}}$$
Table of contents

Ideal gas is characterized by the following properties:

• It consists of point-like particles of the same mass $$m$$.

• There are no molecular forces between the particles.

• Particle collisions are elastic.

• The velocity direction of the particles is disordered.

• The particle's velocity is given by the Maxwell distribution.

• The gas is described by the ideal gas equation:

## Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

### Exercise #1: Tire Pressure after a Journey

Before starting your journey, you check the tire pressure of your car. A pressure gauge shows a pressure of $$3 \, \text{bar}$$ at a temperature of 20°C. After the journey, you find that the tire has warmed up to 60°C.

Assume that the tire has not expanded, and the air inside the tire can be considered as an ideal gas. What is the pressure inside the tire?

#### Solution to Exercise #1

Use the ideal gas to set up equations for the initial and final states of the tire. Since the tire does not expand according to the exercise, the volume $$V$$ remains the same before and after the journey.

Before the journey, the pressure $$\mathit{\Pi}_{\text{before}} = 3 \, \text{bar}$$ and the temperature $$T_{\text{before}} = 20 \,^{\circ} \text{C}$$. So, the ideal gas equation for this state is: 1 $\mathit{\Pi}_{\text{before}} \, V ~=~ n \, R \, T_{\text{before}}$

After the journey, the pressure has changed to an unknown value $$\mathit{\Pi}_{\text{after}}$$, and the temperature has increased to $$T_{\text{after}} = 60 \,^{\circ} \text{C}$$. Therefore, the ideal gas equation for the tire state after the journey is: 2 $\mathit{\Pi}_{\text{after}} \, V ~=~ n \, R \, T_{\text{after}}$

The volume $$V$$, as well as the amount of substance $$n$$ and the gas constant $$R$$, are all constants. Therefore, bring them to the right side of the equation and the variable pressure and temperature to the left side. Then, equations 1 and 2 transform to: 3 $\frac{\mathit{\Pi}_{\text{before}}}{T_{\text{before}}} ~=~ \frac{n \, R}{V}$ 4 $\frac{\mathit{\Pi}_{\text{after}}}{T_{\text{after}}} ~=~ \frac{n \, R}{V}$

Now, the same constants $$\frac{n \, R}{V}$$ are on the right side in both equations. This means you can equate 3 and 4 to get rid of the constants: 5 $\frac{\mathit{\Pi}_{\text{after}}}{T_{\text{after}}} ~=~ \frac{\mathit{\Pi}_{\text{before}}}{T_{\text{before}}}$

Perfect! Now you have three known quantities and only one unknown quantity in the equation, namely the desired tire pressure $$\mathit{\Pi}_{\text{after}}$$ after the journey. Simply rearrange 5 with respect to $$\mathit{\Pi}_{\text{after}}$$: 6 $\mathit{\Pi}_{\text{after}} ~=~ \frac{\mathit{\Pi}_{\text{before}}}{T_{\text{before}}} \, T_{\text{after}}$

Now just substitute the given values. But BE CAREFUL! Don't forget to first convert the temperature to Kelvin scale, otherwise the result will be incorrect. Also, the pressure is given in bar ($$1 \, \text{bar} = 10^5 \, \frac{\text N}{\text{m}^2}$$) and not in Pascal ($$1\, \text{Pa}= 1 \, \frac{\text N}{\text{m}^2}$$). With $$3\, \text{bar} = 300 000 \, \text{Pa}$$, as well as $$20\, ^{\circ}\text{C} = 293.15 \, \text{K}$$ and $$60\, ^{\circ}\text{C} = 333.15 \, \text{K}$$, you get the desired tire pressure: 7 $\mathit{\Pi}_{\text{after}} ~=~ \frac{ 300 000 \, \text{Pa} }{ 293.15 \, \text{K} } \, 333.15 \, \text{K} ~=~ 340 000 \, \text{Pa}$

This corresponds to a pressure: $$\mathit{\Pi}_{\text{after}} ~=~ 3.4 \, \text{bar}$$.