Question

During the thermodynamic process for an ideal gas as shown in the figure, which of the options hold true?

• A. Temperature change, $\Delta T=0$
• B. Heat interaction, $\Delta Q=0$
• C. Work interaction, $W<0$
• D. Internal energy change, $\Delta U>0$

Answer 1

The correct option is D Internal energy change, $\Delta U>0$

As $P$ and $V$ are increasing, therefore $T$ will also increase.
[from ideal gas equation, $PV=nRT$]
So, $\Delta T>0$.

Now, change in internal energy,
$\Delta U=\frac{nfR\Delta T}{2}$ will also increase as $\Delta T>0$,
So, $\Delta U>0$.

Also, work done by gas is positive as volume is increasing.
$\Rightarrow W>0$

From the first law of thermodynamics,
$\Delta Q=\Delta U+W$,
$\Delta Q>0$ as $\Delta U>0$ and $W>0$.
Therefore, option $\left(d\right)$ is correct.