Question

A sample of ideal gas goes from state $\text{A}$ to state $\text{B}$ for as shown in the graph. Let $W$ be the work done by gas and $\Delta U$ be the change in internal energy along the path $\text{AB}$. Identify the correct statement.

• A. Both $W$ and $\Delta U$ are positive
• B. $W=0$ and $\Delta U$ is negative
• C. $W$ is negative and $\Delta U$ is positive
• D. Both $W$ and $\Delta U$ are negative

Answer 1

The correct option is C $W$ is negative and $\Delta U$ is positive

Here Volume $V$ is on vertical axis, however the work done by gas $W$ is obtained by area under $P-V$ curve.


$\Rightarrow$ In part $\text{I}$,
Volume increases from $\text{A}$ to $\text{O}$. Thus, work done by gas is $+ve$.

However, for Part $\text{II}$,
Volume decreases from $\text{O}$ to $\text{B}$ and a larger area under the curve suggest a higher $-ve$ work.

Hence, for process $\text{A}\to \text{B}$,
$W_{net}<0$

From $\text{A}$ to $\text{B}$

At constant volume, pressure increases So from ideal gas equation, temperature will also increase.

Since, $T_A<T_B$, So, $\Delta U>0.$

Hence, option $\left(c\right)$ is correct.