Answer
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Hint: Boyle temperature is described for real gases. Real gases do not adhere to the law of ideal gas and the particles of real gases have volumes.
Complete step-by-step solution:
Boyle temperature can be defined as the point in the temperature range in which a real gas starts to behave like an ideal gas at a pressure range. The temperature at which the second coefficient in the expression becomes zero is known as a Boyle temperature. This Boyle temperature balances out the attractive and the repulsive forces that are acting on a gas particle.
We can use the virial equation of state to calculate the Boyle temperature. Boyle temperature can be expressed in the terms of the virial coefficients.
$Z = 1 + \dfrac{B}{{{V_m}}} + ....$
Additional information:
The above graph is plotted between the pressure $p$and compressibility factor$Z$. Boyle temperature is nicely marked in the graph. From this point, the real gas starts to behave like an ideal gas. The compression factor $Z$ is given as,
$Z = \dfrac{{p{V_m}}}{{RT}}$
Here, $Z$is the compressibility factor, ${V_m}$ is the volume,$R$ is the gas constant, $T$ is the
temperature and $p$ is the pressure.
$Z = 1$ for an ideal gas. Real gases show some deviation.
Note: Critical Temperature is different from Boyle's temperature. At the critical temperature, a gas shows non-ideal behavior. Critical temperature is lower than the Boyle temperature. At Boyle temperature a gas starts to behave like an ideal gas and for ideal gas, the compressibility factor is 1, that is, $Z = 1$
Complete step-by-step solution:
Boyle temperature can be defined as the point in the temperature range in which a real gas starts to behave like an ideal gas at a pressure range. The temperature at which the second coefficient in the expression becomes zero is known as a Boyle temperature. This Boyle temperature balances out the attractive and the repulsive forces that are acting on a gas particle.
We can use the virial equation of state to calculate the Boyle temperature. Boyle temperature can be expressed in the terms of the virial coefficients.
$Z = 1 + \dfrac{B}{{{V_m}}} + ....$
Additional information:
![seo images](https://www.vedantu.com/question-sets/c8a61468-1ba5-4bfb-8020-ca45dc45283e9153133005624434474.png)
The above graph is plotted between the pressure $p$and compressibility factor$Z$. Boyle temperature is nicely marked in the graph. From this point, the real gas starts to behave like an ideal gas. The compression factor $Z$ is given as,
$Z = \dfrac{{p{V_m}}}{{RT}}$
Here, $Z$is the compressibility factor, ${V_m}$ is the volume,$R$ is the gas constant, $T$ is the
temperature and $p$ is the pressure.
$Z = 1$ for an ideal gas. Real gases show some deviation.
Note: Critical Temperature is different from Boyle's temperature. At the critical temperature, a gas shows non-ideal behavior. Critical temperature is lower than the Boyle temperature. At Boyle temperature a gas starts to behave like an ideal gas and for ideal gas, the compressibility factor is 1, that is, $Z = 1$
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