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.

	Column - I		Column - 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 on one mole of monatomic ideal gas is as shown in the

PV

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

Correct answer

D. $a - q, b - r, c - s, d - u$

Step-by-step solution

$\left(a\right){W}_{1 \to 2 \to 3} = {W}_{1 \to 2} + {W}_{2 \to 3}$
$= {P}_{0}\left[2{V}_{0} - {V}_{0}\right] + 0 = {P}_{0}{V}_{0}$
${W}_{1 \to 2 \to 3} = {P}_{0}{V}_{0}$
$\begin{cases}= \dfrac{R{T}_{0}}{3} & \left(a \to q\right)\end{cases}$
\left[ $\begin{array}{ll}\text{Process}1 \to 2 ⇒ \text{isobaric} \\ \text{and}2 \to 3 ⇒ \text{isochoric}\end{array}$ \right] $ \left(b\right){U}_{1 \to 2 \to 3} = \dfrac{3}{2}\left[\dfrac{3{P}_{0}}{2} \times 2{V}_{0} - {P}_{0}{V}_{0}\right]$
$\begin{cases}= \dfrac{3}{2} \times 2{P}_{0}{V}_{0} = 3{P}_{0}{V}_{0} = R{T}_{0} & \left(b \to r\right)\end{cases}$

\left(c\right){Q}_{1 \to 2 \to 3} = {U}_{1 \to 2 \to 3} + {W}_{1 \to 2 \to 3} $ $ \begin{cases}= R{T}_{0} + \dfrac{R{T}_{0}}{3} = \dfrac{4R{T}_{0}}{3} & \left(c \to s\right)\end{cases} $ \left(d\right){Q}_{1 \to 2} = n{C}_{P} ∆ T$
$= n\dfrac{5}{2}R\left({T}_{2} - {T}_{1}\right)$
$= \dfrac{5}{2}\left[{P}_{0}2{V}_{0} - {P}_{0}{V}_{0}\right]$
$= \dfrac{5}{2}{P}_{0}{V}_{0} = \dfrac{5}{2}\dfrac{R{T}_{0}}{3}$
$\begin{cases}= \dfrac{5}{6}{R}_{0}{T}_{0} & \left(d \to u\right)\end{cases}$

Practice Thermodynamics on Quantrex Academy →

More from Thermodynamics