JEE Advanced 2018 · Paper 1 · Chemistry Q6

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ChemistryMultiple correct+4 / -2Question 24 of 54

A reversible cyclic process for an ideal gas is shown below. Here, $P$ , $V$ , and $T$ are pressure, volume and temperature, respectively. The thermodynamic parameters $q$ , $w$ , $H$ and $U$ are heat, work, enthalpy and internal energy, respectively.

The correct option(s) is (are)

V-T diagram of a cycle: A (P₁, V₁, T₁) to C (P₂, V₁, T₂) is horizontal at V₁ (isochoric heating); C to B (P₂, V₂, T₁) is a straight line through the origin direction (isobaric at P₂); B to A is vertical at T₁ (isothermal), with V₂ < V₁.

q_{AC} = \Delta U_{BC}

and

w_{AB} = P_2(V_2 - V_1)

w_{BC} = P_2(V_2 - V_1)

and

q_{BC} = \Delta H_{AC}

\Delta H_{CA} < \Delta U_{CA}

and

q_{AC} = \Delta U_{BC}

q_{BC} = \Delta H_{AC}

and

\Delta H_{CA} > \Delta U_{CA}

Answer and worked solution

Official answer: B, C

Read the cycle (here $w$ is work done on the gas):

A → C: constant $V_1$ , heating $T_1\to T_2$ . Isochoric, so $P$ rises from $P_1$ to $P_2$ .
C → B: the line passes through the origin of the $V$ - $T$ plot, so it is isobaric at $P_2$ . It cools to $T_1$ and contracts to $V_2$ .
B → A: constant $T_1$ . Isothermal expansion from $V_2$ to $V_1$ .

Useful equalities.

$q_{AC}=\Delta U_{AC}=nC_v(T_2-T_1)=\Delta U_{BC}$ , since B and A are both at $T_1$ .
$q_{BC}=\Delta H_{BC}=nC_p(T_2-T_1)=\Delta H_{AC}$ .
$w_{BC}=-P_2(V_1-V_2)=P_2(V_2-V_1)$ .

Check the options.

(A) A → B is not at constant $P_2$ , so $w_{AB}\ne P_2(V_2-V_1)$ . False.
(B) Both parts hold. True.
(C) For C → A, $\Delta T<0$ . $\Delta H_{CA}=nC_p\Delta T$ is more negative than $\Delta U_{CA}=nC_v\Delta T$ , so $\Delta H_{CA}<\Delta U_{CA}$ . With $q_{AC}=\Delta U_{BC}$ , true.
(D) $\Delta H_{CA}>\Delta U_{CA}$ is false.

(B), (C)

Topics:First law and enthalpy Calorimetry

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