IIT-JEE 2012 · Paper 2 · Chemistry Q35

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ChemistryMultiple correct+4Question 35 of 60

The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?

P against V: from (P₁, V₁, T₁) an isothermal curve ends at (P₂, V₂, T₂) and a steeper adiabatic curve ends lower at (P₃, V₂, T₃), both at volume V₂.

T_1=T_2

T_3>T_1

w_{\text{isothermal}}>w_{\text{adiabatic}}

\Delta U_{\text{isothermal}}>\Delta U_{\text{adiabatic}}

Answer and worked solution

Official answer: A, C, D

From the figure. Both expansions start at $(P_1,V_1,T_1)$ and end at $V_2$ ; the adiabat ends at lower pressure.

(A) The isothermal end point has $T_2=T_1$ . True.
(B) An adiabatic expansion cools the gas: $T_3<T_1$ . False.
(C) The gas does more work in the isothermal expansion (more area under the curve). With the IUPAC convention used in chemistry ( $w$ = work done on the system, negative for expansion), $w_\text{isothermal}<w_\text{adiabatic}$ , so the statement is false; if $w$ is read as work done by the gas it is true. The official key accepts both readings.
(D) $\Delta U_\text{iso}=0$ and $\Delta U_\text{adia}=nC_V(T_3-T_1)<0$ , so $\Delta U_\text{iso}>\Delta U_\text{adia}$ . True.

(A), (D) (key: ACD or AD)

Topics:First law and enthalpy

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IIT-JEE 2012 · Paper 2, Chemistry question 35