JEE Advanced 2019 · Paper 1 · Physics Q9

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PhysicsMultiple correct+4 / -1Question 9 of 54

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature $(V$ - $T)$ diagram. The correct statement(s) is/are:

[ $R$ is the gas constant]

V-T diagram of a cycle: 1 (T₀, V₀) → 2 (2T₀, 2V₀) along a straight line through the origin; 2 → 3 (T₀, 2V₀) at constant volume 2V₀; 3 → 4 (T₀/2, V₀) along a line through the origin; 4 → 1 at constant volume V₀.

Work done in this thermodynamic cycle

(1\to2\to3\to4\to1)

|W| = \frac12RT_0

The above thermodynamic cycle exhibits only isochoric and adiabatic processes.

The ratio of heat transfer during processes

1\to2

and

2\to3

\left|\dfrac{Q_{1\to2}}{Q_{2\to3}}\right| = \dfrac53

The ratio of heat transfer during processes

1\to2

and

3\to4

\left|\dfrac{Q_{1\to2}}{Q_{3\to4}}\right| = \dfrac12

Answer and worked solution

Official answer: A, C

Reading the cycle. Lines through the origin on a V-T diagram are isobars.

$1\to2$ is isobaric at $P_1=RT_0/V_0$ .
$2\to3$ is isochoric at $2V_0$ .
$3\to4$ is isobaric at $P_3=RT_0/(2V_0)$ .
$4\to1$ is isochoric at $V_0$ .

So (B) is false.

(A) Work.
$W=P_1V_0-P_3V_0=RT_0-\tfrac12RT_0=\tfrac12RT_0.$
True.

Heats (monatomic gas, $C_p=\frac52R$ , $C_v=\frac32R$ ):

$Q_{1\to2}=\frac52R(T_0)$ .
$Q_{2\to3}=\frac32R(-T_0)$ .
$Q_{3\to4}=\frac52R\left(-\frac{T_0}2\right)$ .

(C) $\left|\dfrac{Q_{1\to2}}{Q_{2\to3}}\right|=\dfrac53$ . True.

(D) $\left|\dfrac{Q_{1\to2}}{Q_{3\to4}}\right|=2$ , not $\frac12$ . False.

(A), (C)

Topics:Thermodynamic processes Work done by a gas Specific heats of gases

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JEE Advanced 2019 · Paper 1, Physics question 9