For isothermal expansion of an ideal gas, the correct combination of the thermodynamic parameters will be
- $\Delta U=0, Q=0, W \neq 0$ and $\Delta H \neq 0$
- $\Delta U \neq 0, Q \neq 0, W \neq 0$ and $\Delta H=0$
- $\Delta U=0, Q \neq 0, W=0$ and $\Delta H \neq 0$
- $\Delta U=0, Q \neq 0, W \neq 0$ and $\Delta H=0$
Solution
$$
\Delta T=0
$$
$\therefore$ From $\Delta U=n C_{v} \Delta T$
$$
\begin{array}{l}
\Delta U=0 \text { and, from } \\
\Delta H=n C_{ho} \Delta t=0
\end{array}
$$
From first law of thermodynamics,
$$
\Delta U=Q+W
$$
as
$$
\Delta U=0 \Rightarrow Q \neq 0
$$
and
$$
W \neq 0
$$
$\therefore$ Parameters are
$$
\Delta U=0, Q \neq 0 ; W \neq 0
$$
and $\Delta H=0$ ,
Asked in: JEE-TOPICTESTS-CHEMISTRY