JEE Advanced · Physics · 13. Thermodynamics

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by $T Δ X,$ where $T$ is temperature of the system and $Δ X$ is the infinitesimal change in a thermodynamic quantity $X$ of the system. For a mole of monatomic ideal gas $X = \frac{3}{2} R l n (\frac{T}{T_{A}}) + R l n (\frac{V}{V_{A}}) .$ Here, $R$ is gas constant, $V$ is volume of gas, $T_{A}$ and $V_{A}$ are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.
List I List II
$(a)$ Work done by the system in process $1 \to 2 \to 3$ $(p)$ $\frac{1}{3} R T_{0} l n 2$
$(b)$ Change in internal energy in process $1 \to 2 \to 3$ $(q)$ $\frac{1}{3} R T_{0}$
$(c)$ Heat absorbed by the system in process $1 \to 2 \to 3$ $(r)$ $R T_{0}$
$(d)$ Heat absorbed by the system in process $1 \to 2$ $(s)$ $\frac{4}{3} R T_{0}$
$(t)$ $\frac{1}{3} R T_{0} (3 + l n 2)$
$(u)$ $\frac{5}{6} R T_{0}$

If the process carried out on one mole of monatomic ideal gas is as shown in figure in the $T$ - $V$ -diagram with $P_{0} V_{0} = \frac{1}{3} R T_{0},$ the correct match is,

	List I		List II
$(a)$	Work done by the system in process $1 \to 2 \to 3$	$(p)$	$\frac{1}{3} R T_{0} l n 2$
$(b)$	Change in internal energy in process $1 \to 2 \to 3$	$(q)$	$\frac{1}{3} R T_{0}$
$(c)$	Heat absorbed by the system in process $1 \to 2 \to 3$	$(r)$	$R T_{0}$
$(d)$	Heat absorbed by the system in process $1 \to 2$	$(s)$	$\frac{4}{3} R T_{0}$
		$(t)$	$\frac{1}{3} R T_{0} (3 + l n 2)$
		$(u)$	$\frac{5}{6} R T_{0}$

A $a - q, b - r, c - p, d - u$
B $a - s, b - r, c - q, d - t$
C $a - q, b - r, c - s, d - u$
D $a - p, b - r, c - t, d - p$

Verified Solution

Answer & Solution

Correct Answer

(D) $a - p, b - r, c - t, d - p$

Step-by-step Solution

Detailed explanation

1

2

process is isothermal and

2

3

process is isochoric.
(a) $\quad W_{1 \rightarrow 2}=n R T \ln \frac{V_f}{F_i}=1 \times R \frac{T_0}{3} \ln \frac{V_2}{V_1}=\frac{R T_0}{3} \ln$ $\frac{2 V_0}{V_0}=\frac{R T_0}{3} \ln 2$

W_{2 \to 3} = 0

(isochoric process)

W_{1 \to 2 \to 3} = W_{1 \to 2} + W_{2 \to 3} = \frac{R T_{0}}{3} \ln 2 (a \to P)

(b) ∆ U = \frac{f}{2} n R (T_{f} - T_{i})

∆ U_{1 \to 2 \to 3} = \frac{3}{2} [1 \times R (T_{0} - \frac{T_{0}}{3})]

\begin{matrix} = R T_{0} & (b \to R) \end{matrix}

(c) Q_{1 \to 2 \to 3} = ∆ U_{1 \to 2 \to 3} + W_{1 \to 2 \to 3}

(First law of thermodynamics)

= R T_{0} + \frac{R T_{0}}{3} \ln 2

\begin{matrix} = \frac{R T_{0}}{3} [3 + \ln 2] & (c \to t) \end{matrix}

(d) Q_{1 \to 2} = ∆ U_{1 \to 2} + W_{1 \to 2}

\begin{matrix} = 0 + \frac{R T_{0}}{3} \ln 2 = \frac{R T_{0}}{3} \ln 2 & (d \to p) \end{matrix}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

List I describes four systems, each with two particles

A

and

B

in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in

m s^{- 1}

) at time

t = \frac{π}{3} s

	List-I		List-II
(I)	$A$ and $B$ are moving on a horizontal circle of radius $1 m$ with uniform angular speed $ω = 1 rad s^{- 1}$ . The initial angular positions of $A$ and $B$ at time $t = 0$ are $θ = 0$ and $θ = \frac{π}{2}$ respectively.	(P)	$\frac{\sqrt{3} + 1}{2}$
(II)	Projectiles $A$ and $B$ are fired (in the same vertical plane) at $t = 0$ and $t = 0.1 s$ respectively, with the same speed $v = \frac{5 π}{\sqrt{2}} m s^{- 1}$ and at $45 °$ from the horizontal plane. The initial separation between $A$ and $B$ is large enough so that they do not collide, $(g = 10 m s^{- 2})$ .	(Q)	$\frac{(\sqrt{3} - 1)}{\sqrt{2}}$
(III)	Two harmonic oscillators $A$ and $B$ moving in the $x$ direction according to $x_{A} = x_{0} \sin \frac{t}{t_{0}}$ and $x_{B} = x_{0} \sin (\frac{t}{t_{0}} + \frac{π}{2})$ respectively, starting from $t = 0$ . Take $x_{0} = 1 m, t_{0} = 1 s$ .	(R)	$\sqrt{10}$
(IV)	Particle $A$ is rotating in a horizontal circular path of radius $1 m$ on the $x y$ plane, with constant angular speed $ω = 1 rad s^{- 1}$ . Particle $B$ is moving up at a constant speed $3 m s^{- 1}$ in the vertical direction as shown in the figure. (Ignore gravity.)	(S)	$\sqrt{2}$
		(T)	$\sqrt{25 π^{2} + 1}$

Which one of the following options is correct?

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