Question 161: Thermodynamics & KTG

Question

Heat given to the processes is positive. Match Column I with Column II:

Column I
Column II

(A)
JK
(P)
$$\Delta W>0$$

(B)
KL
(Q)
$$\Delta \mathrm{Q}0$$

Accepted Answer

$$ \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W}, \mathrm{~W}=\int \mathrm{P} d \mathrm{~V} $$ Along JK : V is constant, P is decreasing and $\mathrm{P} \propto \mathrm{T}, \mathrm{T}$ should also decrease. As $\mathrm{PV}=n \mathrm{RT}$. $$ \begin{aligned} herefore & & \mathrm{W} & =P \Delta V=0 \ & & \Delta \mathrm{U} & =\Delta \mathrm{Q}0, \Delta \mathrm{U}=n \mathrm{C}_{\mathrm{V}} \Delta \mathrm{~T}>0 ext { and } \ & \mathrm{Q}=m c \Delta T>0 . \end{aligned} $$ $ herefore$ Volume increases, $\Delta \mathrm{W}>0$ $\Rightarrow$ Along LM : MV = constant, $\Delta \mathrm{V}=0$ [isochoric process] both P increases and T increases. $$ herefore W=0, \Delta U>0 ext { and } Q>0 \quad[\Delta Q=\Delta U] $$ $\Rightarrow$ Along $\mathrm{MJ}: \mathrm{V}$ is decreasing, $ herefore \Delta \mathrm{W}<0$ T is also decreasing, $\Delta \mathrm{V}<0$ and $\Delta \mathrm{Q}<0$ P increases.

Answer

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{~S}) ; \mathrm{C} \rightarrow(\mathrm{Q}) ; \mathrm{D} \rightarrow( \mathrm{R})]$$

Answer

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{P}) ; \mathrm{C} \rightarrow(\mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{Q})]$$

Answer

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{P}, \mathrm{~S}) ; \mathrm{C} \rightarrow(\mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{Q}, \mathrm{R})] $$

Answer

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{S}) ; \mathrm{C} \rightarrow(\mathrm{S}) ; \mathrm{D} \rightarrow( \mathrm{R})]$$

Thermodynamics & KTG

Heat given to the processes is positive. Match Column I with Column II: Column I Column II (A) JK (P) $\Delta W>0 (B) KL (Q) \Delta \mathrm{Q}<0 (C) LM (R) \Delta \mathrm{W}<0 (D) MJ (S) \Delta Q>0

Choose the correct answer: