IIT-JEE 2010 · Paper 1 · Physics Q69

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PhysicsMultiple correct+3Question 69 of 84

One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in the figure. Its pressure at A is $P_0$ . Choose the correct option(s) from the following

V-T diagram of cycle ABCA: A at (T₀, V₀); A→B vertical up to (T₀, 4V₀); B→C along a straight line through the origin down to C at volume V₀; C→A horizontal at V₀ back to A.

Internal energies at A and B are the same

Work done by the gas in process AB is

P_0V_0\ln4

Pressure at C is

\dfrac{P_0}4

Temperature at C is

\dfrac{T_0}4

Answer and worked solution

Official answer: A, B, C, D

From the V-T diagram.

A → B is at constant $T_0$ (isothermal), from $V_0$ to $4V_0$ .
B → C lies on a line through the origin, $V\propto T$ : isobaric.
C → A is at constant volume $V_0$ .
(A) $U$ depends only on $T$ , and $T_A=T_B$ . True.
(B) $W_{AB}=RT_0\ln\frac{4V_0}{V_0}=P_0V_0\ln4$ . True.
(C) $P_B=\frac{P_0}4$ (Boyle), and B → C is isobaric, so $P_C=\frac{P_0}4$ . True.
(D) At C, $V=V_0$ and $P=\frac{P_0}4$ , so $T_C=\frac{T_0}4$ . True.

(A), (B), (C), (D)

Topics:Thermodynamic processes Work done by a gas

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IIT-JEE 2010 · Paper 1, Physics question 69