Question

According to the first law of thermodynamics, $\Delta U=q+w$. In special cases the statement can be expressed in different ways. Which of the following is not a correct expression?

• A. At constant temperature: $q=-w$
• B. When no work is done: $△U=q$
• C. In gaseous system: $\Delta U=q+P\Delta V$
• D. When work is done by the system: $\Delta U=q+w$

Answer 1

Answer: (D)

When work is done by the system: $\Delta U=q+w$

At constant temperature, $\Delta T=0$

$\therefore \Delta U=n{C}_v\Delta T\therefore \Delta U=q+w=0q=-w$

When work done is zero, $w=0$

$\therefore \Delta U=q+w\therefore \Delta U=q$

In gaseous system, $w=P\Delta V$

$\therefore \Delta U=q+w\therefore \Delta U=q+P\Delta V$

When work is done by the system, w is negative.

$\therefore \Delta U=q-w$