Question

One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the PV diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I. Given Process DA is isothermal.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{A) Process A}\to \text{B}& \text{p) Internal energy decreases.}\\ \text{B) Process B}\to \text{C}& \text{q) Internal energy increases.}\\ \text{C) Process C}\to \text{D}& \text{r) Heat is lost}\\ \text{D) Process D}\to \text{A}& \text{s) Heat is gained.}\\ & \text{t) Work is done one the gas.}\end{array}$

• A. $A\to p,r,t;B\to p,r;C\to q,s;D\to r,t$
• B. $A\to p,r;B\to p,r,t;C\to q,s;D\to r,t$
• C. $A\to p,r,t;B\to q,s;C\to p,r;D\to r,t$
• D. $A\to p,r,t;B\to p,r;C\to r,t;D\to q,s$

Answer 1

The correct option is A $A\to p,r,t;B\to p,r;C\to q,s;D\to r,t$

for (A) : AB is Isobaric compression
W is -Ve; $\Delta U$ is -Ve; $\Delta Q$ is -Ve

for (B) : C is Isochoric
W = 0; $\Delta U$ is -Ve; $\Delta Q$ is -Ve

for (C) : CD is Isobaric expansion
W is +Ve; $\Delta U$ is +Ve; $\Delta Q$ is +Ve

for (D) : DA is Isothermal
W is -Ve; $\Delta U$ = 0; $\Delta Q$ is -Ve