A) \[{{V}_{1}}={{V}_{2}};{{V}_{3}}={{V}_{4}}\text{ and }{{V}_{2}}>{{V}_{3}}\]
B) \[{{V}_{1}}={{V}_{2}};{{V}_{3}}={{V}_{4}}\text{ and }{{V}_{2}}<{{V}_{3}}\]
C) \[{{V}_{1}}={{V}_{2}}={{V}_{3}}={{V}_{4}}\]
D) \[{{V}_{4}}>{{V}_{3}}>{{V}_{2}}>{{V}_{1}}\]
Correct Answer: A
Solution :
[a] From ideal gas equation \[PV=\text{ }\mu \text{RT}\] \[\Rightarrow \]Slope of P.F curve \[\frac{P}{T}=\frac{\mu R}{V}\] \[\Rightarrow \]\[Slope\propto \frac{1}{V}\] It means line of smaller slope represent greater volume of gas. For the given problem figure Point 1 and 2 are on the same line so they will represent same volume i.e. \[{{V}_{1}}={{V}_{2}}\] Similarly point 3 and 4 are on the same line so they will represent same volume i.e. \[{{V}_{3}}={{V}_{4}}\] Also slope of line 1-2 is less than 3-4. Hence \[({{V}_{1}}={{V}_{2}})>\left( {{V}_{3}}={{V}_{4}} \right)\]
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