In a given process, for an ideal gas, $\Delta W=0$ and $\Delta Q < 0$. Then, for the gas,
In a given process, for an ideal gas, $\Delta W=0$ and $\Delta Q < 0$. Then, for the gas,
the temperature will decrease
the volume will increase
the pressure will remain constant
the temperature will increase
Solution
In the given process, for an ideal gas
$\Delta W=0$ and $\Delta Q < 0$
By lst law of thermodynamics,
$\Delta Q=\Delta W+\Delta U$
where, $\Delta U=$ internal energy of gas.
$\begin{aligned} & \Rightarrow \quad \Delta Q=\Delta U \\ & \text { and also } \Delta U < 0 .\end{aligned}$
As the internal energy depends on temperature.
Therefore, if change in internal energy is negative, it mean temperature of gas is decreased.