According to the ideal gas law, if the pressure inside a sealed, rigid container is doubled, what will happen to the temperature of the gas?

• A. The temperature will decrease by a factor of four.
• B. The temperature will double.
• C. The temperature will remain the same.
• D. The temperature will be halved.

Answer: (B)

Verified step by step guidance

1. Recall the ideal gas law, which is given by the equation \(P \times V = n \times R \times T\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.
2. Since the container is sealed and rigid, both the volume \(V\) and the number of moles \(n\) of gas remain constant throughout the process.
3. Rewrite the ideal gas law to relate pressure and temperature directly: \(\frac{P}{T} = \frac{n \times R}{V}\), which means \(\frac{P}{T}\) is a constant when \(n\) and \(V\) are fixed.
4. If the pressure \(P\) is doubled, then to keep the ratio \(\frac{P}{T}\) constant, the temperature \(T\) must also double.
5. Therefore, the temperature of the gas will increase proportionally with the pressure, meaning the temperature will double when the pressure is doubled in a sealed, rigid container.