Question

When the pressure that a gas exerts

Answer 1

Hello!

The answer is:

When the pressure that a gas exerts  on a sealed container changes from

22.5 psi to 19.86 psi, the  temperature changes from 110°C to

65.9°C.

Why?

To calculate which is the last pressure, we need to use Gay-Lussac's law.

The Gay-Lussac's Law states that when the volume is kept constant, the temperature (absolute temperature) and the pressure are proportional.

The Gay-Lussac's equation states that:

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

We are given the following information:

We need to remember that since the temperatures are given in Celsius degrees, we need to convert it to Kelvin (absolute temperature) before use the equation, so:

$P_1=22.5Psi\\T_1=110\°C=110\°C+273.15=383.15K\\T_1=65.9\°C=65\°C+273.15=338.15K$

Now, calculating we have:

$\frac{P_1}{T_1}*(T_2)=P_2\\\\P_2=\frac{P_1}{T_1}*(T_2)=\frac{22.5Psi}{383.15}*338.15=19.86Psi$

Hence, the final pressure is equal to 19.86 Psi.

Have a nice day!

Answer 2

Answer:

The final pressure at 65.9°C is 19.91 psi.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law.

This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$  (at constant Volume)

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=22.5 psi\\T_1=110^oC=383.15 K\\P_2=?\\T_2=65.9^oC=339.05 K$

Putting values in above equation, we get:

$\frac{22.5 psi}{383.15 K}=\frac{p_2}{339.05 K}$

$P_2=\frac{22.5 psi}{383.15 K}\times 339.05 K=19.91 psi$

The final pressure at 65.9°C is 19.91 psi.