In a closed container, the temperature is doubled. What will happen to the pressure?

The question is referring to Gay-Lussac's Law in thermodynamics, which states that the pressure of a gas is directly proportional to its absolute temperature, provided the volume is kept constant.

Here are the steps to understand what happens when the temperature is doubled:

1. Initially, let's assume the pressure of the gas in the container is P1 and the temperature is T1.

2. According to the problem, the temperature is doubled. So, the new temperature, T2, is 2*T1.

3. According to Gay-Lussac's Law, the initial pressure and temperature (P1 and T1) and the final pressure and temperature (P2 and T2) have the following relationship: P1/T1 = P2/T2.

4. Substituting T2 = 2T1 in the equation, we get P1/T1 = P2/(2T1).

5. Simplifying this equation gives us P2 = 2*P1.

So, if the temperature is doubled in a closed container, the pressure will also double, assuming the volume remains constant.

The question is referring to Gay-Lussac's Law in thermodynamics, which states that the pressure of a gas is directly proportional to its absolute temperature, provided the volume is kept constant.

Here are the steps to understand what happens when the temperature is doubled:

1. Initially, let's assume the pressure of the gas in the container is P1 and the temperature is T1.

2. According to the problem, the temperature is doubled. So, the new temperature, T2, is 2*T1.

3. According to Gay-Lussac's Law, the initial pressure and temperature (P1 and T1) and the final pressure and temperature (P2 and T2) have the following relationship: P1/T1 = P2/T2.

4. Substituting T2 = 2T1 in the equation, we get P1/T1 = P2/(2T1).

5. Simplifying this equation gives us P2 = 2*P1.

So, if the temperature is doubled in a closed container, the pressure will also double, assuming the volume remains constant.