Question

An ideal gas undergoes an isothermal expansion along a path AB, adiabatic expansion along BC, isobaric compression along CD, isothermal compression along DE, and adiabatic compression along EA, as shown in the figure. The work done by the gas along the process BC is 10 J. The change in the internal energy along process EA is 16 J. The absolute value of the change in the internal energy along the process CD is ______ J.

Answer 1

6 J

Answer 2

For a complete cycle, the net change in internal energy is zero. Hence, summing the changes:

$$\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CD} + \Delta U_{DE} + \Delta U_{EA} = 0$$

Given:

• Isothermal processes (AB and DE): $\Delta U_{AB} = \Delta U_{DE} = 0$
• Adiabatic process BC: since $W_{BC} = 10\text{ J}$ (work done by the gas) and $Q=0$, we have: $$\Delta U_{BC} = Q - W_{BC} = 0 - 10 = -10\text{ J}$$
• In process EA (adiabatic compression), $\Delta U_{EA} = 16\text{ J}$.

Now, substituting into the cycle equation:

$$0 + (-10) + \Delta U_{CD} + 0 + 16 = 0$$ $$\Delta U_{CD} + 6 = 0 \quad \Rightarrow \quad \Delta U_{CD} = -6\text{ J}$$

The problem asks for the absolute value of $\Delta U_{CD}$. Thus,

$$|\Delta U_{CD}| = 6\text{ J}$$