JEE Main 2016 · 9 Apr · Shift 1 · Chemistry Q32

ChemistrySingle correct+4 / -1Question 32 of 90

At very high pressures, the compressibility factor of one mole of a gas is given by :

\dfrac{pb}{RT}

1+\dfrac{pb}{RT}

1-\dfrac{pb}{RT}

1-\dfrac b{(VRT)}

Answer and worked solution

Official answer: B

At very high pressure the volume term dominates in $p(V-b)=RT$ , so $Z=\frac{pV}{RT}=1+\frac{pb}{RT}$ .

JEE Main 2016 · 9 Apr · Shift 1, Chemistry question 32