Question

which step will decrease the pressure of a gas inside a closed cubical container? increasing the number of moles of gas decreasing the volume of the container increasing the speed of the gas particles decreasing the temperature inside the container

Explanation:

According to the ideal - gas law $PV = nRT$ (where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant, and $T$ is temperature). In a closed container, volume $V$ is constant. When $T$ decreases, and $V$ and $n$ are constant, $P$ will decrease as $P=\frac{nR}{V}T$. Increasing the number of moles $n$ increases $P$. Decreasing volume $V$ increases $P$. Increasing the speed of gas particles is equivalent to increasing temperature which increases $P$.

Answer:

decreasing the temperature inside the container