There is no change in internal energy of an ideal gas when it undergoes

$Assertion :$ Thermodynamic processes in nature are irreversible.
$Reason :$ Dissipative effects cannot be eliminated.

$A$ container of volume $1 \ m^3$ is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at $300 \ K$. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and the gas expands to occupy the whole volume of the container. Its temperature now would be ..... $K$.

In a thermodynamic process of uncontrolled change satisfying the equation $Q=W=0$,where $Q$ is heat supplied and $W$ is work done,the process is:

$A$ container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure $P$ and temperature $T$,whereas the other part is completely evacuated. If the valve is suddenly opened,the pressure and temperature of the gas will be

$A$ thermally insulated container is divided into two parts by a screen. In one part,an ideal gas is filled at pressure $P$ and temperature $T$. The second part is a vacuum. If a small hole is now created in the screen,the temperature of the gas will: