At low pressures, van der Waals equation is written as \[\left( \text{P+}\dfrac{\text{a}}{{{\text{V}}^{\text{2}}}} \right)\text{V = RT}\]. The compressibility factor is then equal to:
a.) \[\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\]
b.) \[\left( \text{1 - }\dfrac{\text{RTV}}{\text{a}} \right)\]
c.) \[\left( \text{1 + }\dfrac{\text{a}}{\text{RTV}} \right)\]
d.) \[\left( \text{1 + }\dfrac{\text{RTV}}{\text{a}} \right)\]
Answer
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Hint: Van Der Waals equation is a variation of the ideal gas equation, i.e., it talks about how real gases behave. This is done by adding correction factors ‘a’ and ‘b’ with volume and pressure respectively. Compressibility factor on the whole is a factor to check the deviation from the ideal gas equation.
Complete step by step answer:
A real gas behaves differently from an ideal gas. This can be mathematically expressed by putting correction terms ‘a’ and ‘b’ in the ideal gas equation.
Ideal gas equation – PV = nRT
Van der Waals equation –
According to the question, we’ve been given Van der Waals equation
Van der Wall’s = \[\left( \text{P+}\dfrac{\text{a}{{\text{n}}^{\text{2}}}}{{{\text{V}}^{\text{2}}}} \right)\left( \text{V - nb} \right)\text{ = nRT}\]
Let the number of moles, n = 1.
This equation can now be written as –
\[\left( \text{P+}\dfrac{\text{a}}{{{\text{V}}^{\text{2}}}} \right)\left( \text{V - b} \right)\text{ = RT}\]
Since the pressure is low, we can ignore the term ‘b’. Therefore, the equation becomes –
\[\left( \text{P+}\dfrac{\text{a}}{{{\text{V}}^{\text{2}}}} \right)\text{V = RT}\]
Now, let us solve this and rearrange the terms –
\[\text{PV + }\dfrac{\text{a}}{\text{V}}\text{ = RT}\]
\[\text{PV = RT - }\dfrac{\text{a}}{\text{V}}\]
Therefore, we get –
\[\text{PV = RT}\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\cdots \cdots \cdots \text{(i)}\]
Compressibility factor is a factor which is used to calculate deviation from ideal gas behaviour. It is indicated by the term ‘Z’.
Equation for a real gas can be written as –
\[\text{PV = ZnRT}\] \[\cdots \cdots \cdots \text{(ii)}\]
Comparing (i) and (ii), we can say that compressibility factor is equal to –
Z = \[\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\]
Therefore, the answer is – option (a) – \[\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\].
Additional Information:
Compressibility factor can also be expressed as a ratio of the product of pressure and volume divided by nRT.
Note: In Van der Waals equation there are two correction terms, ‘a’ and ‘b’.
A correction term is added to pressure (a), because at higher pressure, molecules of a gas do not strike the walls of the container with full pressure because of the attractive forces between them.
In case of volume, we subtract a corrective quantity (b), because molecules are not able to occupy all the volume available due to some repulsive forces acting between them.
Complete step by step answer:
A real gas behaves differently from an ideal gas. This can be mathematically expressed by putting correction terms ‘a’ and ‘b’ in the ideal gas equation.
Ideal gas equation – PV = nRT
Van der Waals equation –
According to the question, we’ve been given Van der Waals equation
Van der Wall’s = \[\left( \text{P+}\dfrac{\text{a}{{\text{n}}^{\text{2}}}}{{{\text{V}}^{\text{2}}}} \right)\left( \text{V - nb} \right)\text{ = nRT}\]
Let the number of moles, n = 1.
This equation can now be written as –
\[\left( \text{P+}\dfrac{\text{a}}{{{\text{V}}^{\text{2}}}} \right)\left( \text{V - b} \right)\text{ = RT}\]
Since the pressure is low, we can ignore the term ‘b’. Therefore, the equation becomes –
\[\left( \text{P+}\dfrac{\text{a}}{{{\text{V}}^{\text{2}}}} \right)\text{V = RT}\]
Now, let us solve this and rearrange the terms –
\[\text{PV + }\dfrac{\text{a}}{\text{V}}\text{ = RT}\]
\[\text{PV = RT - }\dfrac{\text{a}}{\text{V}}\]
Therefore, we get –
\[\text{PV = RT}\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\cdots \cdots \cdots \text{(i)}\]
Compressibility factor is a factor which is used to calculate deviation from ideal gas behaviour. It is indicated by the term ‘Z’.
Equation for a real gas can be written as –
\[\text{PV = ZnRT}\] \[\cdots \cdots \cdots \text{(ii)}\]
Comparing (i) and (ii), we can say that compressibility factor is equal to –
Z = \[\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\]
Therefore, the answer is – option (a) – \[\left( \text{1 - }\dfrac{\text{a}}{\text{RTV}} \right)\].
Additional Information:
Compressibility factor can also be expressed as a ratio of the product of pressure and volume divided by nRT.
Note: In Van der Waals equation there are two correction terms, ‘a’ and ‘b’.
A correction term is added to pressure (a), because at higher pressure, molecules of a gas do not strike the walls of the container with full pressure because of the attractive forces between them.
In case of volume, we subtract a corrective quantity (b), because molecules are not able to occupy all the volume available due to some repulsive forces acting between them.
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