1

### JEE Main 2017 (Online) 9th April Morning Slot

An ideal gas undergoes isothermal expansion at constant pressure. During the process :
A
enthalpy increases but entropy decreases.
B
enthalpy remains constant but entropy increases.
C
enthalpy decreases but entropy increases.
D
Both enthalpy and entropy remain constant.
2

### JEE Main 2017 (Online) 9th April Morning Slot

A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :
A
10 J of the work will be done by the gas.
B
6 J of the work will be done by the gas.
C
10 J of the work will be done by the surrounding on gas.
D
6 J of the work will be done by the surrounding on gas.
3

### JEE Main 2018 (Offline)

The combustion of benzene(l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25oC; heat of combustion (in kJ mol–1) of benzene at constant pressure will be
(R = 8.314 JK–1 mol–1)
A
–3267.6
B
4152.6
C
–452.46
D
3260

## Explanation

Formula of Heat of combination is

$\Delta H = \Delta u\,\, + \,\,\Delta ng\,\,RT$

Where, $\Delta H$ $=$ Heat of combination at constant pressure

$\Delta u\, =$ Heat at constant volume

$\Delta {n_g}$ change in number of moles for gaseous molecule.

R = gas constant

T = Temperature.

The Required reaction

C6H6(l) + ${{15} \over 2}$ O2(g) $\to$ 6CO2(g) + 3H2O(l)

Here O2 and CO2 are gaseous molecules so to calculate $\Delta {n_g}$ we only consider those.

$\therefore\,\,\,$ $\Delta {n_g}$ $=$ 6 $-$ ${{15} \over 2}$ = $-$ ${3 \over 2}$

Given $\Delta$u = $-$ 3263.9 kJ mol$-$1 R = 8.314 JK$-$ mol$-$1

R = 8.314 JK$-$1 mol$-$1

= 8.314 $\times$ 10$-$3 kJ K$-$1 mol$-$1

T = 25o C

= 25 + 273 K

= 298 K

So, $\Delta H$ = $-$ 3263.9 + $\left( { - {3 \over 2}} \right)$ $\times$ 8.314 $\times$ 10$-$3 $\times$ 298

= $-$ 3267.6 kJ mol$-$1
4

### JEE Main 2018 (Offline)

Which of the following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction? A
A and D
B
A and B
C
B and C
D
C and D

## Explanation

We know from thermodynamics,

$\Delta$Go = $\Delta$Ho $-$ T$\Delta$So

$\Rightarrow \,\,\,\, - RT\ln K = \Delta {H^o} - T\Delta {S^o}$

$\Rightarrow \,\,\,\,\ln K = - {{\Delta {H^o}} \over R} \times {1 \over T} + {{\Delta {S^o}} \over R}.......\left( 1 \right)$

We know, equation of straight line is

$y = mx + c\,\,\,.........\,\,\,\left( 2 \right)$

by comparing (1) and (2) we get,

slope (m) = $-$ ${{\Delta {H^o}} \over R}$ and intercept (c) = ${{\Delta {S^o}} \over R}$

For exothermic reaction $\Delta {H^o} = - Ve$

$\Rightarrow \,\,\,\, - \Delta {H^o} = + ve$

$\Rightarrow \,\,\,\,{{ - \Delta {H^o}} \over R} = + ve$

$\therefore\,\,\,$ Slope is positive, so, graph A and B is possible.

Now if entropy $\left( {\Delta {S^o}} \right)$ is positive then intercept $\left( {{{\Delta {S^o}} \over R}} \right)$ is positive. It is represented by graph A.

If entropy $\left( {\Delta {S^o}} \right)$ is negative then intercept $\left( {{{\Delta {S^o}} \over R}} \right)$ is negative. It is represented by graph B. 