Question

$nk$mol of a monochromatic ideal gas is taken quasi-statically from state A to state C along the straight line shown in the given figure. Alternatively, the same gas is shown in the taken quasi-statically from A to C along the path ABC.

Match the column (i) with column (ii) on the basis of the given information.

Column (I)	Column (II)
i. The heat $\Delta H$ added to the gas along the straight line path AC	a.$2P_A{V}_A$
ii. Change in $\Delta U$ of the gas along the straight line path AC	b. $\frac{13}{2}{P}_A{V}_A$
iii. The work done by the gas along the path ABC	c. $6P_A{V}_A$
iv. The heat $\Delta H$ added to the gas along the path ABC	d. $\frac{9}{2}{P}_A{V}_A$

• A. $i\to c;ii\to d;iii\to a;iv\to b$
• B. $i\to a;ii\to c;iii\to b;iv\to d$
• C. $i\to b;ii\to a;iii\to c;iv\to d$
• D. $i\to d;ii\to a;iii\to b;iv\to c$

Answer 1

The correct option is A $i\to c;ii\to d;iii\to a;iv\to b$

(i). Heat, $\Delta H$ added to the gas along the straight line path $AC$.
(a) Work $\Delta W_{AC}$ done by the gas along the straght line path $AC$
$\Delta W_{AC}=\left({\int }_A^{C}PdV\right)$ i.e., area under the path.
$\Delta W_{AC}$=$P_A{V}_A+\frac{1}{2}{P}_A{V}_A=\frac{3}{2}{P}_A{V}_A$
(b). Increase of internal energy is $\Delta U_{AC}$
and, $\Delta U_{AC}=\frac{3}{2}nRT\left(T_C-T_A\right)=\frac{3}{2}\left(P_C{V}_C-P_A{V}_A\right)=\frac{9}{2}{P}_A{V}_A$
Hence, heat added,
$\Delta H_{AC}=\Delta W_{AC}+\Delta U_{AC}=6P_A{V}_A$

(ii). From above,
$\Delta U_{AC}=\frac{9}{2}{P}_A{V}_A$

(iii).
Work done is$\Delta W_{ABC}=\Delta W_{AB}+\Delta W_{BC}=0+P_C(V_C-V_A)=2P_A{V}_A$

(iv).
Heat $\Delta H$ added to the gas along $ABC$,
$\Delta H_{ABC}=\Delta H_{AB}+\Delta H_{BC}=\frac{9}{2}{P}_A{V}_A+2P_A{V}_A=\frac{13}{2}{P}_A{V}_A$