Answer the following by appropriately matching the lists based on...

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 $T\DeltaX,$ where $T$ is temperature of the system and $\DeltaX$ is the infinitesimal change in a thermodynamic quantity $X$ of the system. For a mole of monatomic ideal gas $X = \dfrac{3}{2}Rln\left(\dfrac{T}{{T}_{A}}\right) + Rln\left(\dfrac{V}{{V}_{A}}\right) .$ 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 $1 \to 2 \to 3$	$( p )$	$\dfrac{1}{3}R{T}_{0}\ln 2$
$( b )$	Change in internal energy in process $1 \to 2 \to 3$	$( q )$	$\dfrac{1}{3}R{T}_{0}$
$( c )$	Heat absorbed by the system in process $1 \to 2 \to 3$	$( r )$	$R{T}_{0}$
$( d )$	Heat absorbed by the system in process $1 \to 2$	$( s )$	$\dfrac{4}{3}R{T}_{0}$
		$( t )$	$\dfrac{1}{3}R{T}_{0}\left(3 + \ln 2\right)$
		$( u )$	$\dfrac{5}{6}R{T}_{0}$

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} = \dfrac{1}{3}R{T}_{0},

the correct match is,
Answer the following by appropriately matching the lists based on the information given in the paragraph. In a

Options

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 $a - p, b - r, c - t, d - p$

Correct answer

D. $a - p, b - r, c - t, d - p$

Step-by-step solution

-

process is isothermal and

-

process is isochoric. <br/>\left(a\right){W}_{1 \to 2} = nRTln\dfrac{{V}_{f}}{{F}_{i}} = 1 \times R\dfrac{{T}_{0}}{3}ln \dfrac{{V}_{2}}{{V}_{1}} = \dfrac{R{T}_{0}}{3}ln \dfrac{2{V}_{0}}{{V}_{0}} = \dfrac{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} = \dfrac{R{T}_{0}}{3}\ln 2\left(a \to P\right)

\left(b\right) ∆ U = \dfrac{f}{2}n

R\left({T}_{f} - {T}_{i}\right)

∆ {U}_{1 \to 2 \to 3} = \dfrac{3}{2}\left[1 \times R\left({T}_{0} - \dfrac{{T}_{0}}{3}\right)\right]

\begin{cases}= R{T}_{0} & \left(b \to R\right)\end{cases}

\left(c\right){Q}_{1 \to 2 \to 3} = ∆ {U}_{1 \to 2 \to 3} + {W}_{1 \to 2 \to 3}

(First law of thermodynamics) <br/>

= R{T}_{0} + \dfrac{R{T}_{0}}{3}\ln 2

<br/>

\begin{cases}= \dfrac{R{T}_{0}}{3}\left[3 + \ln 2\right] & \left(c \to t\right)\end{cases}

<br/>\left(d\right){Q}_{1 \to 2} = ∆ {U}_{1 \to 2} + {W}_{1 \to 2}

\begin{cases}= 0 + \dfrac{R{T}_{0}}{3}\ln 2 = \dfrac{R{T}_{0}}{3}\ln 2 & \left(d \to p\right)\end{cases}

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