A container with rigid walls filled with a sample of ideal gas. The absolute temperature of the gas is doubled. What happens to the pressure of the gas?

the pressure is directly proportional to the absolute temperature

God 0

Ideal gas

To understand what happens to the pressure of the gas when the absolute temperature is doubled, we can refer to the ideal gas law. The ideal gas law states that the pressure (P) of an ideal gas is directly proportional to the absolute temperature (T) of the gas, when the volume (V) and the number of gas particles (n) are constant.

Mathematically, the ideal gas law can be represented as:

PV = nRT

Where:
P = pressure
V = volume
n = number of gas particles
R = ideal gas constant
T = absolute temperature

Since we are considering a case where the volume and the number of gas particles are constant (as described in the question), we can rewrite the ideal gas law as:

P1/T1 = P2/T2

Where:
P1 = initial pressure
T1 = initial absolute temperature
P2 = final pressure (when the temperature is doubled)
T2 = final absolute temperature (when the temperature is doubled)

In this case, we are doubling the absolute temperature, so T2 = 2T1. Substituting these values into the equation, we get:

P1/T1 = P2/(2T1)

To isolate P2, we can cross-multiply the equation:

P1 * 2T1 = P2 * T1

Now, we can cancel out T1 from both sides:

P1 * 2 = P2

This equation shows that the final pressure (P2) will be doubled (2P1) when the absolute temperature (T) is doubled while the volume and the number of gas particles remain constant.

In summary, when the absolute temperature of an ideal gas is doubled in a container with rigid walls and constant volume and number of gas particles, the pressure of the gas also doubles.