Question

Fig. shows an indicator diagram. During path $1-2-3$, $100\text{cal}$ are given to the system and $40\text{cal}$ worth work is done. During path $1-4-3$, the work done is $10\text{cal}$.

• A. Heat given to the system during path $1-4-3$ is $70\text{cal}$
• B. If the system is brought from $3$ to $1$ along straight line path $3-1$, work done is worth $25\text{cal}$
• C. Along straight line path $3-1$, the heat ejected by the system is $85\text{cal}$
• D. The internal energy of the system in state $3$ is $140\text{cal}$ above that in state $1$.

Answer 1

Answer: (A), (B), (C)

Along straight line path $3-1$, the heat ejected by the system is $85\text{cal}$

Work in path $1-3$ is mean of that for $1-2-3$ and $1-4-3$ (considering areas).

for process $1-2-3$
$\Delta Q=\Delta U+W$
$100=\Delta U+40\Rightarrow \Delta U=60cal$
For process $1-4-3$
$\Delta Q=\Delta U+W$
$\Delta Q=60+10=70cal$
Option A is correct.

For process $3-1$
$\Delta U=-60cal$

Geometrically we can say that area under $3-1$ is half of area under $1-2-3\&1-4-3$
$W=-\frac{40+10}{2}=-25cal$
here -ve sign is because process $3-1$ is compressive in nature and work is done on the system.

Now $\Delta Q=\Delta U+W=-25-60=-85cal$