1

### JEE Main 2019 (Online) 9th January Morning Slot

Which one of the following statements regarding Henry's law is not correct ?
A
Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids
B
Different gases have different KH (Henry's law constant) values at the same temperature.
C
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
D
The value of KH increases with increase of temperature and KH is function of the nature of the gas

## Explanation

From Henry's law we know,

Pgas = KH xg

Where, Pgas = Pressure of undissolved gas

xg = Mole fraction of gas dissolved into the liquid.

KH = Henry's Carnot.

When pressure is constant then,

xg $\propto$ ${1 \over {K{}_H}}$

So. when KH is high then xg or solubility of gas is lower in the liquid.

So, option (A) is wrong.
2

### JEE Main 2019 (Online) 9th January Morning Slot

Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 and T2 (T1 < T2). The correct graphical depiction of the dependence of work done (w) on the final volume (V) is :
A B C D ## Explanation

Work done in isothermal process,

W = $-$ nRT ln${{{V_2}} \over {{V_1}}}$

$\therefore$   $\left| W \right|$ = nRT ln${{{V_2}} \over {{V_1}}}$

$\Rightarrow$   $\left| W \right|$ = nRT (lnV2 $-$ lnV1)

$\Rightarrow$   $\left| W \right|$ = nRT lnV2 $-$ nRT lnV1

given that $V$2 $=$ $V$

$\therefore$   $\left| W \right|$ = nRT lnV $-$ nRT lnV1

By comparing with the straight line equation,

y = mx + C

We get slope is nRT and intercept $-$ nRT lnV1 in $\left| W \right|$ and lnV graph.

As T2 > T1   So,

Slope nRT2 > nRT1

and intercept

$-$ nRT2 lnV1 < $-$ nRT1 lnV1

So, we can say

(1)   slope of T2 line is more then T1

(2)  intercept of T1 line is less negative than T2 line, and intercept of T1 can't be positive, it can be 0 or less than 0 as $-$ nRT1lnV1 always $\le$ 0
3

### JEE Main 2019 (Online) 9th January Evening Slot

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is :

(Specific heat of water liquid and water vapour are 4.2 kJ K$-$1 kg$-$1 and 2.0 kJ K$-$1 kg$-$1; heat of liquid fusion and vapourisation of water are 334 kJ$-$1 and 2491 kJ kg$-$1, respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)
A
7.90 kJ kg$-$1 K$-$1
B
2.64 kJ kg$-$1 K$-$1
C
8.49 kJ kg$-$1 K$-$1
D
9.26 kJ kg$-$1 K$-$1

## Explanation $\Delta {S_1} = {{\Delta {H_{fusion}}} \over {273}} = {{334} \over {273}} = 1.22$

$\Delta {S_2} = 4.2\ell N\left( {{{363} \over {273}}} \right) = 1.31$

$\Delta {S_3} = {{\Delta {H_{vap}}} \over {373}} = {{2491} \over {373}} = 6.67$

$\Delta {S_4} = 2.0\ell n\left( {{{383} \over {373}}} \right) = 0.05$

$\Delta {S_{total}} = 9.26\,$ kJ Kg$-$1 K$-$1
4

### JEE Main 2019 (Online) 10th January Morning Slot

A process has $\Delta$H = 200 J mol–1 and $\Delta$S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous:
A
4 K
B
20 K
C
5 K
D
12 K

## Explanation

$\Delta$G = $\Delta$H $-$ T$\Delta$S

T = ${{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K$