
At temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as $\text{p = }\dfrac{\text{RT}}{\text{V - b}}$ at T. Here, b is the van der Waals constant. Which gas will exhibit the steepest increase in the plot Z (compression factor) vs p?
A. Ne
B. Ar
C. Xe
D. Kr
Answer
581.4k+ views
Hint: For this problem, we have to study the relationship of compressibility factor and the size of each element. So, those elements which have greater size will show the increase in the slope of the graph.
Complete step-by-step Answer:
- In the given question, we have to explain which element will show the steepest increase in the graph between the compressibility factor and pressure.
- Now, it is given in the question that the gas is ideal and for the ideal gas the relationship between pressure, volume and temperature will be:
$\text{PV = nRT}$
- But the real gas shows different behaviour from that of the ideal gas such that there is an increase in the pressure whereas the decrease in the volume takes place.
- So, the real gas equation will become $\left( \text{P + }\dfrac{\text{a}}{{{\text{V}}^{2}}} \right)(\text{V - b)}\,\text{ = RT}$
- Now, when the pressure becomes high the value of pressure becomes only P and the equation of the real can be written as:
$\text{P}(\text{V - b)}\,\text{ = RT}$
$\Rightarrow \dfrac{\text{PV}}{\text{RT}}\text{ - }\dfrac{\text{Pb}}{\text{RT}}\text{ = 1}$
$\Rightarrow \text{z = 1 + }\dfrac{\text{Pb}}{\text{RT}}$ …. (1)
So, from equation (1) we can see that the size of the molecule of the gas that is represented by 'b' is directly proportional to the compressibility factor.
- Now, we know that among neon, argon, xenon and krypton the size of the xenon is large because we know that down the group the size of the atom increases.
Therefore, option C is the correct answer.
Note: The compressibility factor helps determine the behaviour of the real gas and how much deviation the real gas shows from the ideal gas. Also, the compressibility factor for the ideal gas is 1.
Complete step-by-step Answer:
- In the given question, we have to explain which element will show the steepest increase in the graph between the compressibility factor and pressure.
- Now, it is given in the question that the gas is ideal and for the ideal gas the relationship between pressure, volume and temperature will be:
$\text{PV = nRT}$
- But the real gas shows different behaviour from that of the ideal gas such that there is an increase in the pressure whereas the decrease in the volume takes place.
- So, the real gas equation will become $\left( \text{P + }\dfrac{\text{a}}{{{\text{V}}^{2}}} \right)(\text{V - b)}\,\text{ = RT}$
- Now, when the pressure becomes high the value of pressure becomes only P and the equation of the real can be written as:
$\text{P}(\text{V - b)}\,\text{ = RT}$
$\Rightarrow \dfrac{\text{PV}}{\text{RT}}\text{ - }\dfrac{\text{Pb}}{\text{RT}}\text{ = 1}$
$\Rightarrow \text{z = 1 + }\dfrac{\text{Pb}}{\text{RT}}$ …. (1)
So, from equation (1) we can see that the size of the molecule of the gas that is represented by 'b' is directly proportional to the compressibility factor.
- Now, we know that among neon, argon, xenon and krypton the size of the xenon is large because we know that down the group the size of the atom increases.
Therefore, option C is the correct answer.
Note: The compressibility factor helps determine the behaviour of the real gas and how much deviation the real gas shows from the ideal gas. Also, the compressibility factor for the ideal gas is 1.
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