Answer
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Hint: Recall the meaning of compressibility factor, critical temperature, pressure, and volume. Think about how all of these are related and expressed in terms of the Vander Waals constants.
Complete answer:
Here, ${{Z}_{c}}$ is the compressibility factor which shows the deviation of the behaviour of a real gas from the behaviour of an ideal gas. The Vander Waals constants ‘a’ and ‘b’ are used to calculate the value of ${{Z}_{c}}$ at critical temperature, pressure, and volume.
From here onwards we will refer to ‘a’ and ‘b’ as Van Der Waals constants, ${{T}_{c}}$ as critical temperature, ${{P}_{c}}$ as critical pressure, and ${{V}_{c}}$ as critical volume.
- Critical temperature (${{T}_{c}}$): It is the temperature above which the gas cannot be liquefied at any pressure. It’s relation to the Vander Waals constants is:
\[{{T}_{c}}=\dfrac{8a}{27Rb}\]
- Critical pressure (${{P}_{c}}$): It is the pressure required to liquefy a real gas at its critical temperature. It’s relation to the Vander Waals constants is:
\[{{P}_{c}}=\dfrac{a}{27{{b}^{2}}}\]
- Critical volume (${{V}_{c}}$): It is the volume of one mole of the real gas liquefied at the critical temperature. It’s relation to the Vander Waals constants is:
\[{{V}_{c}}=3b\]
Now, to calculate the compressibility factor, we need to put these values of ${{T}_{c}}$, ${{P}_{c}}$ and ${{V}_{c}}$ into the formula calculating ${{Z}_{c}}$. The formula for compressibility factor is:
\[{{Z}_{c}}=\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}\]
Now, putting the values, we get:
\[{{Z}_{c}}=\dfrac{\dfrac{a}{27{{b}^{2}}}\times 3b}{R\times \dfrac{8a}{27Rb}}\]
\[{{Z}_{c}}=\dfrac{\dfrac{a}{9b}}{\dfrac{8a}{27b}}\]
\[{{Z}_{c}}=\dfrac{a}{9b}\times \dfrac{27b}{8a}\]
\[{{Z}_{c}}=\dfrac{3}{8}\]
Hence, the answer is ‘A. 3/8’
Note: The compressibility factor in the critical state of any real gas is known to be 3/8.
The relation between the critical states and the Vander Waals constant is derived using Van Der Waals real gas equation:
\[[P+\dfrac{a{{n}^{2}}}{{{V}^{2}}}]+[V-nb]=nRT\]
Complete answer:
Here, ${{Z}_{c}}$ is the compressibility factor which shows the deviation of the behaviour of a real gas from the behaviour of an ideal gas. The Vander Waals constants ‘a’ and ‘b’ are used to calculate the value of ${{Z}_{c}}$ at critical temperature, pressure, and volume.
From here onwards we will refer to ‘a’ and ‘b’ as Van Der Waals constants, ${{T}_{c}}$ as critical temperature, ${{P}_{c}}$ as critical pressure, and ${{V}_{c}}$ as critical volume.
- Critical temperature (${{T}_{c}}$): It is the temperature above which the gas cannot be liquefied at any pressure. It’s relation to the Vander Waals constants is:
\[{{T}_{c}}=\dfrac{8a}{27Rb}\]
- Critical pressure (${{P}_{c}}$): It is the pressure required to liquefy a real gas at its critical temperature. It’s relation to the Vander Waals constants is:
\[{{P}_{c}}=\dfrac{a}{27{{b}^{2}}}\]
- Critical volume (${{V}_{c}}$): It is the volume of one mole of the real gas liquefied at the critical temperature. It’s relation to the Vander Waals constants is:
\[{{V}_{c}}=3b\]
Now, to calculate the compressibility factor, we need to put these values of ${{T}_{c}}$, ${{P}_{c}}$ and ${{V}_{c}}$ into the formula calculating ${{Z}_{c}}$. The formula for compressibility factor is:
\[{{Z}_{c}}=\dfrac{{{P}_{c}}{{V}_{c}}}{R{{T}_{c}}}\]
Now, putting the values, we get:
\[{{Z}_{c}}=\dfrac{\dfrac{a}{27{{b}^{2}}}\times 3b}{R\times \dfrac{8a}{27Rb}}\]
\[{{Z}_{c}}=\dfrac{\dfrac{a}{9b}}{\dfrac{8a}{27b}}\]
\[{{Z}_{c}}=\dfrac{a}{9b}\times \dfrac{27b}{8a}\]
\[{{Z}_{c}}=\dfrac{3}{8}\]
Hence, the answer is ‘A. 3/8’
Note: The compressibility factor in the critical state of any real gas is known to be 3/8.
The relation between the critical states and the Vander Waals constant is derived using Van Der Waals real gas equation:
\[[P+\dfrac{a{{n}^{2}}}{{{V}^{2}}}]+[V-nb]=nRT\]
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