Question

An ideal gas undergoes a four step cycle as shown in the $P–V$ diagram below. During this cycle, heat is absorbed by the gas in

• A. steps $1\text{and}2$
• B. steps$1\text{and}3$
• C. steps$1\text{and}4$
• D. steps $2\text{and}4$

Answer 1

The correct option is C steps$1\text{and}4$


Process $1:$
$\text{P = Constant}$
$\text{Volume = increasing}$
$V\propto T$
Temperature will increase
$\Delta T=+ve$
$\Delta Q=n{C}_p\Delta T=+ve$

Process $2:$
$\text{V = Constant and pressure = decreasing}$
$P\propto T$ Temperature = decreasing
$\Delta Q=n{C}_v\Delta T\Rightarrow \Delta T=-ve\Rightarrow \Delta Q=-ve$

Process $3:$
$P=\text{Constant}$ and $\text{Vol = decreasing}$
$V\propto T\Rightarrow \text{T = decreasing}$
$\Delta T=-ve\Rightarrow \Delta Q=n{C}_p\Delta T=-ve$

Process $4:$
$V=\text{Constant}\text{and}P=\text{increasing}$
$P\propto T\Rightarrow \text{T = increasing}$
$\Delta T=+ve\Rightarrow \Delta Q=n{C}_p\Delta T=+ve$
So, heat is absorbed by the gas in steps $1$ and $4$.