1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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.

Questions Asked from Thermodynamics

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