
The compressibility factor for a real gas at high pressure is:
A. $1+\dfrac{RT}{pb}$
B. 1
C. $1+\dfrac{pb}{RT}$
D. $1-\dfrac{pb}{RT}$
Answer
590.1k+ views
Hint: The compressibility factor (Z), which is useful for modifying the ideal gas to check the behaviour of real gases. It is actually a correction factor that describes the deviation of a real gas from ideal gas. Calculate the ratio of \[\dfrac{PV}{RT}\] from the Van Der Waal equation of the real gases and then you will get the answer.
Complete step by step answer:
- Firstly, we will write Van Der Waal equation of real gases:
\[\left( P+\dfrac{a}{{{v}^{2}}} \right)\left( v-b \right)=RT\]
- This above equation is the Van Der Waal equation for one mole of a gas.
- It is found that at high pressure,
$P>>\dfrac{a}{{{V}^{2}}}$
Here, $\dfrac{a}{{{v}^{2}}}$ will be negligible.
So, the equation transforms in the form of:
\[\begin{align}
& P(V-b)=RT \\
& PV-Pb=RT \\
& PV=RT+Pb \\
\end{align}\]
Now, we will get:
\[\dfrac{PV}{RT}=1+\dfrac{Pb}{RT}\]
- Compressibility factor is represented by Z, we can write:
\[Z=\dfrac{Pb}{RT}\]
Here, we will substitute value of=\[\]\[\dfrac{Pb}{RT}\]
We will get the expression of compressibility factor:
\[Z=1+\dfrac{Pb}{RT}\]
- Hence, we can conclude that the correct option is (C), that is The compressibility factor for a real gas at high pressure is $1+\dfrac{pb}{RT}$
So, the correct answer is “Option C”.
Additional Information:
- It is also found that is the ratio of the actual molar volume of the gas to the calculated molar volume of the gas at the same pressure and temperature. We can also write it as:
\[Z=\dfrac{{{V}_{real}}}{{{V}_{ideal}}}\]
- Compressibility factor is also called as gas deviation factor
- Compressibility factor depends on the particular gas as well as pressure and temperature.
Note: - It is also found that Compressibility factor has a value of one for an ideal gas.
- We should not get confused in between the terms Compressibility factor and compressibility. As, compressibility of a material is the measure of the relative volume change of a solid or fluid in response to a pressure change. Whereas, Compressibility factor describes the deviation of real gas from ideal gas behaviour.
Complete step by step answer:
- Firstly, we will write Van Der Waal equation of real gases:
\[\left( P+\dfrac{a}{{{v}^{2}}} \right)\left( v-b \right)=RT\]
- This above equation is the Van Der Waal equation for one mole of a gas.
- It is found that at high pressure,
$P>>\dfrac{a}{{{V}^{2}}}$
Here, $\dfrac{a}{{{v}^{2}}}$ will be negligible.
So, the equation transforms in the form of:
\[\begin{align}
& P(V-b)=RT \\
& PV-Pb=RT \\
& PV=RT+Pb \\
\end{align}\]
Now, we will get:
\[\dfrac{PV}{RT}=1+\dfrac{Pb}{RT}\]
- Compressibility factor is represented by Z, we can write:
\[Z=\dfrac{Pb}{RT}\]
Here, we will substitute value of=\[\]\[\dfrac{Pb}{RT}\]
We will get the expression of compressibility factor:
\[Z=1+\dfrac{Pb}{RT}\]
- Hence, we can conclude that the correct option is (C), that is The compressibility factor for a real gas at high pressure is $1+\dfrac{pb}{RT}$
So, the correct answer is “Option C”.
Additional Information:
- It is also found that is the ratio of the actual molar volume of the gas to the calculated molar volume of the gas at the same pressure and temperature. We can also write it as:
\[Z=\dfrac{{{V}_{real}}}{{{V}_{ideal}}}\]
- Compressibility factor is also called as gas deviation factor
- Compressibility factor depends on the particular gas as well as pressure and temperature.
Note: - It is also found that Compressibility factor has a value of one for an ideal gas.
- We should not get confused in between the terms Compressibility factor and compressibility. As, compressibility of a material is the measure of the relative volume change of a solid or fluid in response to a pressure change. Whereas, Compressibility factor describes the deviation of real gas from ideal gas behaviour.
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