The value of ${{Z}_{c}}$ at ${{T}_{c}}$, ${{P}_{c}}$ and ${{V}_{c}}$ is:
A. 3/8
B. 4/8
C. 1
D. 0
Answer
611.4k+ views
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:
\[[P+\dfrac{a{{n}^{2}}}{{{V}^{2}}}]+[V-nb]=nRT\]
Recently Updated Pages
Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain the structure of megasporangium class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

How many chromosomes are found in human ovum a 46 b class 12 biology CBSE

The diagram of the section of a maize grain is given class 12 biology CBSE

