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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Chemistry
Chapter
States of Matter

JEE Advanced 2015 · Paper 2 · Chemistry · States of Matter

One mole of a monoatomic real gas satisfied the equation p(V - b) =...

One mole of a monoatomic real gas satisfied the equation p(V−b)=RTp\left(V - b\right) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given byOne mole of a monoatomic real gas satisfied the equation p(V - b) = RT where b is a constant. The relationship ofOne mole of a monoatomic real gas satisfied the equation p(V - b) = RT where b is a constant. The relationship ofOne mole of a monoatomic real gas satisfied the equation p(V - b) = RT where b is a constant. The relationship ofOne mole of a monoatomic real gas satisfied the equation p(V - b) = RT where b is a constant. The relationship of

Options

  1. AWhat is the normal range of glomerular filtration rate (GFR) in a healthy individual?
  2. BWhat is the normal range of glomerular filtration rate (GFR) in a healthy individual?
  3. CWhat is the normal range of glomerular filtration rate (GFR) in a healthy individual?
  4. DWhat is the normal range of glomerular filtration rate (GFR) in a healthy individual?

Correct answer

C. What is the normal range of glomerular filtration rate (GFR) in a healthy individual?

Step-by-step solution

P(V−b)=RTP\left(V - b\right) = RT
→PV−Pb=RT\to PV - Pb = RT
→PVRT=PbRT+1\to \dfrac{PV}{RT} = \dfrac{Pb}{RT} + 1
→Z=1+PbRT\to Z = 1 + \dfrac{Pb}{RT}
Hence, Z>1Z > 1 at all pressures.
This means, repulsive tendencies will be dominant when interatomic distance are small. This means, interatomic potential is never negative but becomes positive at small interatomic distances.
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