1

### 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
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### 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
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### 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 = ${{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K$
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### JEE Main 2019 (Online) 10th January Morning Slot

The values of Kp/Kc for the following reactions at 300 K are, respectively : (At 300 K, RT = 24.62 dm3 atm mol–1)

N2(g) + O2(g)  $\rightleftharpoons$ 2 NO(g)

N2O4(g)  $\rightleftharpoons$   2 NO(g)

N2(g) + 3H2(g) $\rightleftharpoons$ 2 NH3(g)
A
24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2 1.65 $\times$ 10–3 dm–6 atm–2 mol2
B
1,4.1 $\times$ 10–2 dm–3 atm–1 mol, 606 dm6 atm2 mol–2
C
1,24.62 dm3 atm mol–1 , 1.65 $\times$ 10–3 dm–6 atm –2 mol2
D
1, 24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2

## Explanation 