Question

A process cycle ABCA consisting of isothermal expansion AB, isobaric compression BC and adiabatic compression CA, for 1 mole of ideal gas is shown below. (Given: $T_A=T_B=400K,\lambda =1.5;(2{)}^{\frac{1}{3}}=1.26)$


$\begin{array}{cccccc}& ColumnI& & ColumnII& & ColumnIII\\ \left(I\right)& \Delta Q=-247.5R& \left(i\right)& \Delta U=165R& \left(P\right)& \Delta W=165R\\ \left(II\right)& \Delta Q=247.5R& \left(ii\right)& \Delta U=-165R& \left(Q\right)& \Delta W=400R\left(ln2\right)\\ \left(III\right)& \Delta Q=400R\left(ln2\right)& \left(iii\right)& \Delta U=82.5R& \left(R\right)& \Delta W=-165R\\ \left(IV\right)& \Delta Q=zero& \left(iv\right)& \Delta U=zero& \left(S\right)& \Delta W=-82.5R\end{array}$

Which of the following options is correct for process $A\to B$?

• A. (II)(iii)(P)
• B. (I)(ii)(S)
• C. (III)(iv)(Q)
• D. (III)(iii)(P)

Answer 1

The correct option is C (III)(iv)(Q)

$A\to B$ (isothermal)
$\Delta Q=\Delta U+\Delta W$
$\Delta U=0(\therefore \Delta T=0)$
$\Delta Q=\Delta W=nRTln\frac{v_2}{v_1}=400R\left(ln2\right)$
$B\to C$ isobaric
$T_c=317.48K$ (using $P^{1-\gamma }{T}^{\gamma }=constant$ for C to A)
Now,
$\Delta Q=n{C}_p\Delta T=-247.5R$
$\Delta U=n{C}_v\Delta T=-165R$
$\Delta W=nR\Delta T=-82.5R$
Process $C\to A$ (adiabatic)
$\Delta Q=0$
$\Delta W=-\Delta U$
$\Delta U=n{C}_v\Delta T=165R$
$\Delta W=-165R$