# Ideal Gas Law

## What is an Ideal Gas?

An ideal gas is a hypothetical gas! It doesn’t exist in actuality, yet is expected to exist to streamline computations. It additionally creates a reference point in connection to which conduct of different gases can be contemplated.

An ideal gas is characterized as a gas made out of arbitrarily moving particles as all gases do, the main contrast being that for an ideal gas, when its particles slam into one another, their crashes are thought to be consummately versatile ,which means no vitality of both of these particles is squandered.

In all actuality, be that as it may, when real gas particles slam into one another, a portion of their vitality is squandered in changing ways and conquering grating. Nonetheless, at STP (characterized underneath) conditions most characteristic gases act simply like an ideal gas exposed to sensible limitations.

## Ideal Gas Equation

The ideal gas equation is as follows:

PV = nRT

where
P = pressure of the gas (atm)
V = volume of the gas (L)
n = Number of Moles
T = Absolute temperature (Kelvin)
R = Ideal Gas constant otherwise called Boltzmann Constant = 0.082057 L atm K-1 mol-1.

Important conversion factors:

Pressure: 1 atm = 101.3 kPa = 760 mmHg
Volume: 1000 ml = 1 L = 1000 cm3
Temperature: K = °C + 273

Example 1 – 6.2 liters of an ideal gas is contained at 3.0 atm and 37°C. How many moles of this gas are present?

Step 1: Write down your given information:

V = 6.2 L
P = 3.0 atm
T = 37°C

Step 2: Convert units

Because the units of the gas constant are given using atmospheres, moles, and Kelvin, it’s important to make sure you convert values given in other temperature or pressure scales. For this problem, convert °C temperature to K using the equation:

T = °C + 273
T = 37°C + 273
T = 310 K

Step 3: Plug in the variables into the appropriate equation. Rearrange the equation to solve for moles:

n = PV / RT
n = (3.0 atm x 6.2 L) / (0.08 L atm /mol K x 310 K)
n = 0.75 mol

There are 0.75 mol of ideal gas present in the system.

Example 2 – 5.0 g of neon is at 256 mm Hg and at a temperature of 35°C. What is the volume?

Step 1: Write down your given information:
P = 256 mmHg
V = ?
m = 5.0 g
R = 0.0820574 L•atm•mol-1K-1
T = 35°C

Step 2: Convert units

We must convert the pressure to atm, temperature to kelvin, and grams of neon to moles of neon: Step 3: Substitute the variables into the appropriate equation. Rearrange the equation to solve for volume:

V = nRT/P

V = [(¼ moles) (0.08 L atm /mol K) (308 K)] / (0.336 atm)

V = 19L

Example 3 – What is a gas’s temperature in Celsius when it has a volume of 25 L, 203 mol, 143.5 atm?

Step 1: Write down your given information:

P = 143.5 atm
V = 25 L
n = 203 mol
R = 0.0820574 L•atm•mol-1 K-1

Step 2: Skip because all units are the appropriate units.

Step 3: Plug in the variables into the appropriate equation.

T = PV/nR

T = [(143.5atm) (25L)] / [(203mol) (0.08206L•atm/Kmol)]

T = 215.4K

Step 4: You are not done. Be sure to read the problem carefully, and answer what they are asking for. In this case, they are asking for temperature in Celsius, so you will need to convert it from K, which are the units you have.

215.4K − 273 = −57.4°C