A thermometer with a liquid-filled bulb at one end, the most commonly used liquid are Mercury, Toluene, Alcohol, Pentane show different readings for temperatures other than the fixed reading because of their different expansion properties.

Whereas 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. The variables that describe the behaviour of a given quantity (mass) of gas are pressure, volume, and temperature \((P, V\), and \(T)\) (where \(T=t+273.15\); \(t\) is the temperature in \({ }^{\circ} \mathrm{C}\) ).

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,

\(
P \propto 1 / V
\)
or
\(PV =\) Constant
Where \(P\) is the pressure and \(V\) is the volume.

The above relationship is known as Boyle’s law, after Robert Boyle (1627–1691), the English Chemist who discovered it.

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, 

\(V \propto T\)
or
\(V / T=\) Constant
where \(T\) is the Temperature and \(V\) is the volume.

The above relationship is known as Charles’ law, after French scientist Jacques Charles (1747–1823).

Low-density gases obey these laws, which may be combined into a single relationship. This relationship is known as the ideal gas law.

Notice that since \(P V=\) constant and \(V / T=\) constant for a given quantity of gas,
then \(P V / T\) should also be a constant. 
It can be written in a more general form that applies not just to a given quantity of a single gas but to any quantity of any low-density gas and is known as the ideal gas equation:
\(
\begin{aligned}
& \frac{P V}{T}=\mu R \\
& \text { or } P V=\mu R T \dots(11.2)
\end{aligned}
\)

where \(\mu\) is the number of moles in the sample of gas and \(R\) is called the universal gas constant:
\(
R=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
\)

Absolute Temperature

Thermodynamic temperature is another name for absolute temperature. The thermodynamic energy of a system is lowest at this temperature. Absolute temperature equals zero Kelvin or \(-273.15^{\circ} \mathrm{C}\), commonly known as absolute zero. The velocity of the gas particles stops at an 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.

In Eq. 11.2, we have learnt that the pressure and volume are directly proportional to temperature : \(P V \propto T\). This relationship allows a gas to be used to measure temperature in a constant-volume gas thermometer. Holding the volume of a gas constant, it gives \(P \propto T\). Thus, with a constant-volume gas thermometer, the temperature is read in terms of pressure. A plot of pressure versus temperature gives a straight line in this case, as shown in Flg. 11.2.

However, measurements on real gases deviate from the values predicted by the ideal gas law at low temperatures. But the relationship is linear over a large temperature range, and it looks as though the pressure might reach zero with decreasing temperature if the gas continued to be a gas. The absolute minimum temperature for an ideal gas, therefore, inferred by extrapolating the straight line to the axis, as in Fig. 11.3. This temperature is found to be \(-273.15^{\circ} \mathrm{C}\) and is designated as absolute zero.

Absolute zero is the foundation of the Kelvin temperature scale or absolute scale temperature named after the British scientist Lord Kelvin. On this scale, \(-273.15^{\circ} \mathrm{C}\) is taken as the zero point, that is \(0 \mathrm{~K}\) (Fig. 11.4).

The size of unit in Kelvin and Celsius temperature scales is the same. So, the temperature on these scales are related by
\(
T=t_{\mathrm{C}}+273.15 \dots(11.3)
\)

Example 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 \(273.0 \mathrm{~K}\).
Pressure is \(1.00 \mathrm{~atm}\).
The universal gas constant \((R)\) has a value of \(0.08206 \mathrm{Latm} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\).
The expression for the number of moles is,
\(\mu=m / M\)
where, \(\mu\) 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 \(44.0 \mathrm{~g} \mathrm{~mol}^{-1}\).
So, the value of \(\mu\) can be calculated as,
\(
\begin{aligned}
\mu & =2.34 \mathrm{~g} / 44.0 \mathrm{~g} \mathrm{~mol}^{-1} \\
& =0.0532 \mathrm{~mol}
\end{aligned}
\)
According to the ideal gas equation,
\(
P V=\mu R T
\)
Rearranging the equation,
\(
V=\mu R T / P
\)
Substituting all the values,
\(
\begin{aligned}
V & \left.=[0.0532 \mathrm{~mol})\left(0.08206 \mathrm{Latm} \mathrm{mol}^{-} 1 \mathrm{~K}^{-} 1\right)(273.0 \mathrm{~K})\right] / 1.00 \mathrm{~atm} \\
& =1.19 \mathrm{~L}
\end{aligned}
\)

Example 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 \(273.0 \mathrm{~K}\).
Pressure is \(1.00 \mathrm{~atm}\).
Molar mass of the argon gas is \(39.948 \mathrm{~g} / \mathrm{mol}\).
According to the ideal gas equation,
\(
P V=\mu R T
\)
Rearranging the equation,
\(\mu=P V / R T\)
Substituting all the values in the above equation,
\(
\begin{aligned}
\mu & =[(1.00 \mathrm{~atm})(56.2 \mathrm{~L})] /\left[\left(0.08206 \mathrm{Latm} \mathrm{mol}^{-} 1 \mathrm{~K}^{-} 1\right)(273.0 \mathrm{~K})\right] \\
& =2.50866 \mathrm{~mol}
\end{aligned}
\)
The expression for the number of moles is
\(
\mu=m / M
\)
Rearranging the equation,
\(
m=\mu M
\)
Substituting all the values in the above equation,
\(
\begin{aligned}
m & =(2.50866 \mathrm{~mol}) \times(39.948 \mathrm{~g} / \mathrm{mol}) \\
& =100 \mathrm{~g}
\end{aligned}
\)

Example 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 \(1.95 \mathrm{~atm}\).
The Number of moles is \(0.654\) moles.
According to the ideal gas equation,
\(P V=\mu R T\)
Rearranging the equation,
\(T=P V / \mu R\)
Substituting all the values in the above equation,
\(
\begin{aligned}
T & =[(1.95 \mathrm{~atm}) \times(12.30 \mathrm{~L})] /\left[(0.654 \mathrm{~mol}) \times\left(0.08206 \mathrm{Latm} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\right] \\
& =447 \mathrm{~K}
\end{aligned}
\)

Example 4: 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 \((\mathrm{m})\) of the carbon dioxide is \(5.600 \mathrm{~g}\).
The Volume \((V)\) of the carbon dioxide is \(4.00 \mathrm{~L}\).
The Temperature is \(300 \mathrm{~K}\).
Molar mass of the carbon dioxide is \(44.0 \mathrm{~g} \mathrm{~mol}^{-1}\)
The expression for the number of moles is
\(\mu=m / M\)
Substituting all the values in the above equation,
\(
\begin{aligned}
\mu & =(5.600 \mathrm{~g}) /(44.009 \mathrm{~g} / \mathrm{mol}) \\
& =0.1272467 \mathrm{~mol}
\end{aligned}
\)
According to the ideal gas equation,
\(
P V=\mu R T
\)
Rearranging the equation,
\(
P=\mu R T / V
\)
Substituting all the values in the above equation,
\(
\begin{aligned}
P & =(0.1272467 \mathrm{~mol}) \times\left(0.08206 \mathrm{Latm} \mathrm{mol} \mathrm{m}^{-1} \mathrm{~K}^{-1}\right) \times(300 \mathrm{~K}) /(4.00 \mathrm{~L}) \\
& =0.7831 \mathrm{~atm}
\end{aligned}
\)