Question

You had a closed tank of air at a pressure of 4 atm and temperature of 20 degrees Celsius. When the tank and the air are heated to 40 degrees Celsius, what is the pressure if the volume remains constant?​

Answer 1

Answer:

The pressure will be 4.27 atm.

Explanation:

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, K = Constant

This law indicates that the quotient between pressure and temperature is constant.

This law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of collisions with the walls increases, that is, the pressure increases. That is, the pressure of the gas is directly proportional to its temperature.

In short, when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases.

You want to study two different states, an initial state and a final state. You have a gas that is at a pressure P1 and a temperature T1 at the beginning of the experiment. By varying the temperature to a new value T2, then the pressure will change to P2, and the following will be fulfilled:

$\frac{P1}{T1} =\frac{P2}{T2}$

In this case:

• P1= 4 atm
• T1= 20 C= 293 K (being 0 C= 273 K)
• P2= ?
• T2= 40 C= 313 K

Replacing:

$\frac{4 atm}{293 K} =\frac{P2}{313 K}$

Solving:

$P2= 313 K* \frac{4 atm}{293 K}$

P2= 4.27 atm

The pressure will be 4.27 atm.