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:

Step1: Recall ideal gas law

The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant, and $T$ is temperature.

Step2: Analyze each option

• Option 1: Increasing $n$ (number of moles) while $V$ and $T$ are constant will increase $P$ according to $P=\frac{nRT}{V}$.
• Option 2: Decreasing $V$ while $n$ and $T$ are constant will increase $P$ since $P=\frac{nRT}{V}$.
• Option 3: Increasing the speed of gas particles means increasing the temperature (as temperature is related to the average kinetic energy of gas particles). Increasing $T$ while $n$ and $V$ are constant will increase $P$ according to $P=\frac{nRT}{V}$.
• Option 4: Decreasing $T$ while $n$ and $V$ are constant. From $P=\frac{nRT}{V}$, when $T$ decreases, $P$ will decrease.

Answer:

decreasing the temperature inside the container