# Ideal Gas Equation and Absolute Temperature

A thermometer with a liquid-filled bulb at one end, the most commonly used liquid are Mercury, Toluene, Alcohol, Pentane, Creosote show different readings for temperatures other than the fixed reading because of their different expansion properties. A thermometer that uses a gas, on the other hand, shows the same reading for temperatures. It does not matter which type of gas is used. Experiments show that all gases expand in the same way at low densities.

Pressure (P), volume (V), and temperature (T) where T = t + 273.15 and t is the temperature in °C are the variables that explain the behavior of a given quantity (mass) of gas. The ideal gas law, also known as the universal gas equation, is a state-of-equation for a hypothetical ideal gas. Despite its flaws, the ideal gas law provides a good approximation of the behavior of many gases under a variety of situations. Benoit Paul Émile Clapeyron proposed the ideal gas law in 1834 as a mixture of the empirical Charles’ law, Boyle’s law, Avogadro’s law, and Gay-law. Lussac’s.

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### Ideal Gas Equation

Ideal gas laws are the combination of the observational work of Boyle in the seventeenth century and Charles in the eighteenth century.

**Boyle’s law:** The gas pressure is inversely proportional to the gas volume for a given amount of gas kept at a fixed temperature i.e. at constant temperature the relation between the pressure and volume of a quantity of gas can be written as,

P ∝ 1 / Vor

PV = Constantwhere P is the pressure and V is the volume.

**Charles’ law:** The gas volume is directly proportional to the gas temperature for a given fixed amount of gas kept at a constant pressure i.e. at constant temperature the relation between the volume and temperature of a quantity of gas can be written as,

V ∝ Tor

V / T = Constantwhere T is the Temperature.

These two laws apply to low-density gases and can be grouped into a single relationship. It’s worth noting that,

**PV = Constant**

and

**V/T = Constant**

For a particular quantity of gas, then

**PV/T** should thus be a constant as well.

It can be stated in a more general form that applies to any quantity of any low-density gas, not simply a specific quantity of that gas. This relationship describes the ideal gas law and knows as the **ideal gas equation**.

It can be expressed as,

PV / T = nRor

PV = nRTwhere, n

is the number of moles in the sample of gas and R is the universal gas constant.

Note:The universal gas constant (R) has a value of 8.314 kJ/mole in the SI system.

It can also be stated in a more general form that applies to any quantity of any low-density gas, not simply a specific quantity of that gas.

**Derivation of the Ideal Gas Equation**

Let P is the pressure exerted by the gas, V is the volume f the gas, T is the Temperature.

According to Boyle’s Law,

P ∝ 1/Vor

V ∝ 1/P……(1)According to Charles’ Law,

V ∝ T……..(2)According to Avogadro’s Law,

When P and T are both constant, the volume of a gas is proportional to the number of moles of gas.i.e.

V ∝ n…….(3)Compare equation (1), (2) and (3) as,

V ∝ nT/Por

PV = nRTwhere

Ris the Universal gas constant and it is value of 8.314 J/mol-K

**Absolute Temperature**

Thermodynamic temperatureis another name forabsolute temperature. The thermodynamic energy of a system is lowest at this temperature. Absolute temperature equals zero Kelvin or -273 °C, commonly known asabsolute zero. The velocity of the gas particles stops at absolute zero temperature. This signifies that the particles of the gas really aren’t moving. At absolute zero, the volume of the gas is zero. As a result, the volume of a gas is measured by its absolute zero.

The temperature has a direct relationship with pressure and volume i.e.

**PV ∝ T**

This relationship enables a gas that will be utilized to determine the temperature in a Gas thermometer with a constant volume.

Therefore, at constant volume, the relationship can be written as,** **

**P ∝ T**, and Temperature is read in terms of pressure with a constant-volume gas thermometer.

A straight line emerges from a plot of pressure against temperature.

Observations on real gases differ from the values anticipated by the ideal gas law at low temperatures. However, the relationship is linear over a wide temperature range, and it appears that if the gas remained a gas, the pressure would drop to zero with decreasing temperature. Extrapolating the straight line to the axis yields the absolute minimum temperature for an ideal gas. Absolute zero is defined as a temperature of – 273.15 degrees Celsius. The Kelvin temperature scale, often known as absolute scale temperature, is founded on absolute zero.

On the Kelvin temperature scale, – 273.15 °C is taken as the zero points, that is 0 K. In both the Kelvin and Celsius temperature systems, the unit size is the same. So, the relation between them can be expressed as

**T = t + 273.15**

where t is the temperature in °C

**Sample Problems **

**Problem 1: What is the volume occupied by 2.34 grams of carbon dioxide gas at STP?**

**Solution:**

Given,

Weight (m) of the carbon dioxide is 2.34 grams.

At STP, Temperature is 273.0 K.

Pressure is 1.00 atm.

The universal gas constant (R) has a value of 0.08206 L atm mol¯

^{1}K¯^{1}.The expression for the number of mole is,

n = m/M

where, n is the number of moles, m is the weight and M is the molar mass of the substance.

Molar mass of the carbon dioxide is 44.0 g mol¯

^{1}.So, the value of n can be calculated as,

n = 2.34 g / 44.0 g mol¯

^{1}= 0.0532 mol

According to the ideal gas equation,

PV = nRT

Rearranging the equation,

V = nRT / P

Substituting all the values,

V = [0.0532 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K)] / 1.00 atm

=

1.19 L

**Problem 2: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass of argon in the sample.**

**Solution:**

Given,

Volume (V) of the argon gas is 56.2 liters.

At STP, Temperature is 273.0 K.

Pressure is 1.00 atm.

Molar mass of the argon gas is 39.948 g/mol.

According to the ideal gas equation,

PV = nRT

Rearranging the equation,

n = PV / RT

Subtitituting all the values in the above equation,

n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K)]

= 2.50866 mol

The expression for the number of mole is

n = m/M

Rearranging the equation,

m = nM

Subtitituting all the values in the above equation,

m = (2.50866 mol)×(39.948 g/mol)

=

100 g

**Problem 3:At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?**

**Solution:**

Given,

The Volume (V) of the neon gas is 12.30 liters.

The Pressure is 1.95 atm.

The Number of moles is 0.654 moles.

According to the ideal gas equation,

PV = nRT

Rearranging the equation,

T = PV / nR

Subtitituting all the values in the above equation,

T = [(1.95 atm) ×(12.30 L)] / [(0.654 mol)×(0.08206 L atm mol¯

^{1 }K¯^{1})]=

447 K

**Problem 4: Derive the Ideal Gas Equation?**

**Solution:**

Let P is the pressure exerted by the gas, V is the volume f the gas, T is the Temperature.

According to Boyle’s Law,

P ∝ 1/Vor

V ∝ 1/P……(1)According to Charles’ Law,

V ∝ T……..(2)According to Avogadro’s Law,

When P and T are both constant, the volume of a gas is proportional to the number of moles of gas.i.e.

V ∝ n…….(3)Compare equation (1), (2) and (3) as,

V ∝ nT/Por

PV = nRTwhere

Ris the Universal gas constant and it is value of 8.314 J/mol-K

**Problem 5: 5.600 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of 300 K. When all the solid CO2 becomes gas, what will be the pressure in the container?**

**Solution:**

Given,

The Weight (m) of the carbon dioxide is 5.600 g.

The Volume (V) of the carbon dioxide is 4.00 L.

The Temperature is 300 K.

Molar mass of the carbon dioxide is 44.0 g mol¯1

The expression for the number of mole is

n = m/M

Subtitituting all the values in the above equation,

n = (5.600 g) / (44.009 g/mol)

= 0.1272467 mol

According to the ideal gas equation,

PV = nRT

Rearranging the equation,

P = nRT/V

Subtitituting all the values in the above equation,

P = (0.1272467 mol)× (0.08206 L atm mol¯

^{1 }K¯^{1})× (300 K)/ (4.00 L)=

0.7831 atm