Question

A sample consisting of 1mol of a monoatomic perfect gas $\left(\begin{array}{c}{C}_v=\frac{3}{2}R\end{array}\right)$ is taken through the cycle as shown :

Temperature at points (1), (2) and (3), respectively is :

• A. $273K,546K,273K$
• B. $546K,273K,273K$
• C. $273K,273K,273K$
• D. $546K,546K,273K$

Answer 1

Answer: (A)

$273K,546K,273K$

Given :
$n=1mol$

Consider point 1 :-

$P_1$ = 1 atm (from graph)
$V_1$ = 22.4 L (from graph)

Using $PV=nRT$ we get,

$T_1$ = $\frac{P_1{V}_1}{nR}$
$⟹T_1=273K$.

For point 2 :-
$P_2=P_1$
$V_2=2\times V_1$
Hence $T_2=2\times T_1=546K$.

Consider point 3 :-
$P_1{V}_1=P_3{V}_3$.
It is an isothermal process. Hence $T_3=T_1=273K$
Hence, option A is the right answer.