**JEE Advanced Previous Year Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.**

*Simulator*

**Previous Years JEE Advance Questions**

**[JEE 2009]**

**Sol.**4

(A) nb

(B) $\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}$

$(\mathrm{C})-\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}$

(D) –nb

**[JEE 2009]**

**Sol.**(B)

Attractive force term of real gas is represented by $\frac{\mathrm{n}^{2} \mathrm{a}}{\mathrm{v}^{2}}$

** [JEE 2011]**

**Sol.**7

PV = nRT

0.32 × V = 0.1 × 0.0821 × 273

V = 7 litre

(A) 1.0

(B) 4.5

(C) 1.5

(D) 3.0

**[JEE 2012]**

**Sol.**(C)

If $\mathrm{b}=0 ;\left(\mathrm{P}+\frac{\mathrm{a}}{\mathrm{Vm}^{2}}\right) \times \mathrm{Vm}=\mathrm{RT}$

$\mathrm{PVm}+\frac{\mathrm{a}}{\mathrm{Vm}}=\mathrm{RT}$

$\mathrm{PVm}+\frac{\mathrm{a}}{\mathrm{Vm}}=\mathrm{RT}$

of the form, y = c + mx

here, m = –a

$\mathrm{m}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=\frac{21.6-20.1}{2-3.0}=-1.5$

$\therefore \mathrm{a}=1.5 \mathrm{atm} \mathrm{L}^{2} \mathrm{mol}^{-2}$

**Paragraph for Question 5 & 6**

X and Y are two volatile liquids with molar weights of 10g $\operatorname{mol}^{-1}$ and 40g $\operatorname{mol}^{-1}$ respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours.

(A) 8

(B) 12

(C) 16

(D) 20

**[JEE 2014]**

**Sol.**(C)

(A) Larger mean free path for X as compared to that of Y

(B) Larger mean free path for Y as compared to that of X

(C) Increased collision frequency of Y with the inert gas as compared to that of X with the

inert gas

(D) Increased collision frequency of X with the inert gas as compared to that of Y with the

inert gas

**[JEE 2014]**

**Sol.**(D)

(A)

(B)

(C)

(D)

**[JEE 2015]**

**Sol.**(C)

P(v – b) = RT

Using thermodynamic equation of state, we get

**[JEE – Adv. 2016]**

**Sol.**4

Rate of diffusion $\propto \lambda \times U_{\text {Avg }}$

$\propto \frac{1}{\sqrt{2} \pi \sigma^{2} \mathrm{N}^{*}} \times \mathrm{U}_{\mathrm{Avg}}$

$\propto \frac{\mathrm{U}_{\mathrm{Avg}}}{\sqrt{2} \pi \sigma^{2} \mathrm{N}^{*}}$

$\propto \frac{\mathrm{U}_{\mathrm{Avg}}(\mathrm{kT})}{\sqrt{2 \pi \sigma^{2} \mathrm{P}}}$

Rate of diffusion $\propto \frac{\mathrm{T}^{\frac{3}{2}}}{\mathrm{P}}$

$\frac{\mathrm{r}_{\mathrm{final}}}{\mathrm{r}_{\mathrm{initial}}}=\frac{(4)^{\frac{3}{2}}}{2}$

$\frac{\mathrm{r}_{\text {final }}}{\mathrm{r}_{\text {inital }}}=4$

) of the compartment A after the system attains equilibrium is____.

**[JEE – Adv. 2018]**

**Sol.**-14.6

$2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})$

$\mathrm{p}_{\mathrm{H} 2}$ is the minimum partial pressure of $\mathrm{H}_{2}$ (in bar) needed to prevent the oxidation at 1250 K. The value of ln $\left(\mathrm{p}_{\mathrm{H} 2}\right)$ is ___.

(Given : total pressure = 1 bar, R (universal gas constant) = 8 $\mathrm{JK}^{-1} \mathrm{mol}^{-1}$ , ln(10) = 2.3. Cu(s) and $\mathrm{Cu}_{2} \mathrm{O}(\mathrm{s})$ are mutually immiscible.

At $1250 \mathrm{K}: 2 \mathrm{Cu}(\mathrm{s})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{Cu}_{2} \mathrm{O}(\mathrm{s}) ; \Delta \mathrm{G}^{\theta}=-78,000 \mathrm{J} \mathrm{mol}^{-1}$

$\left.\mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{G}^{\theta}=-1,78,000 \mathrm{J} \mathrm{mol}^{-1} ; \mathrm{G} \text { is the Gibbs energy }\right)$

**[JEE – Adv. 2018]**

**Sol.**2.22