JEE Advanced 2021 Paper 1 Online

An ideal gas undergoes a reversible isothermal expansion from state I to state II followed by a reversible adiabatic expansion from state II to state III. The correct plot(s) representing the changes from state I to state III is (are)

(p : pressure, V : volume, T : temperature, H : enthalpy, S : entropy)

Explanation

From state I to II Reversible isothermal expansion takes place. So, following changes take place,

● pressure decreases.
● Volume increases.
● Temperature remains constant.
● Enthalpy, H remains constant.
● Entropy, S for expansion increases.

So, all options follows the above mentioned conditions, so all graphs are correct for state I and II.

From state II to III Reversible adiabatic expansion takes place. So, following changes take place.

● pressure decreases.
● Volume increases.
● Temperature decreases.
● Enthalpy, H decreases.
● Entropy, S remains constant.
● H increases instead of decreasing, so only option (c) is incorrect.

All other options, i.e., (a), (b) and (d) follows the above mentioned conditions.

Therefore, correct graphical representations are (a), (b) and (d).

JEE Advanced 2020 Paper 1 Offline

MCQ (More than One Correct Answer)

In thermodynamics, the p-V work done is given by

$$w = - \int {dV{p_{ext}}} $$

For a system undergoing a particular process, the work done is

$$w = - \int {dV\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)} $$

This equation is applicable to a

system that satisfies the van der Walls' equation of state

process that is reversible and isothermal

process that is reversible and adiabatic

process that is irreversible and at constant pressure

Explanation

Given, $$w = - \int {{p_{ext}}dV} $$

For 1 mole van der Walls' gas

$$p = \left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)$$

For reversible process, p_ext = p_gas

$$w = - \int {\left( {{{RT} \over {V - b}} - {a \over {{V^2}}}} \right)dV} $$

But, it is not applicable for irreversible process which are carried out very fast. So, work done is calculated assuming final pressure remains constant throughout the process. Thus, statement (a), (b) and (c) correct while statement (d) is incorrect.

JEE Advanced 2019 Paper 1 Offline

MCQ (More than One Correct Answer)

Choose the reaction(s) from the following options, for which the standard enthalpy of reaction of equal to the standard enthalpy of formation.

2C(g) + 3H₂(g) $$ \to $$ C₂H₆(g)

2H₂(g) + O₂(g) $$ \to $$ 2H₂O(l)

$${3 \over 2}$$O₂(g) $$ \to $$ O₃(g)

$${1 \over 8}$$S₈(s) + O₂(g) $$ \to $$ SO₂(g)

Explanation

The standard enthalpy of formation is defined as standard enthalpy change for formation of 1 mole of a substance from its elements, present in their most stable state of aggregation.

$${3 \over 2}$$O₂(g) $$ \to $$ O₃(g);

$${1 \over 8}$$S₈(s) + O₂(g) $$ \to $$ SO₂(g)

In the above two reactions standard enthalpy of reaction is equal to standard enthalpy of formation.

JEE Advanced 2018 Paper 2 Offline

MCQ (More than One Correct Answer)

For a reaction, $$A\,\,\rightleftharpoons\,\,P,$$ the plots of $$\left[ A \right]$$ and $$\left[ P \right]$$ with time at temperature $${T_1}$$ and $${T_2}$$ are given below.

If $${T_2} > {T_1},$$ the correct statement(s) is (are) (Assume $$\Delta {H^ \circ }$$ and $$\Delta {S^ \circ }$$ are independent of temperature and ratio of $$lnK$$ at $${T_1}$$ to $$lnK$$ at $${T_2}$$ is greater than $${{{T_2}} \over {{T_1}}}.$$ Here $$H,$$ $$S,G$$ and $$K$$ are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

$$\Delta {H^ \circ } < 0,\Delta {S^ \circ } < O$$

$$\Delta {G^ \circ } < 0,\Delta {H^ \circ } > 0$$

$$\Delta {G^ \circ } < 0,\Delta {S^ \circ } < 0$$

$$\Delta {G^ \circ } < 0,\Delta {S^ \circ } > 0$$