Respuesta :
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:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
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:
[tex]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[/tex]
Now, calculating we have:
[tex]\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[/tex]
Hence, the final pressure is equal to 19.86 Psi.
Have a nice day!
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,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex] (at constant Volume)
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=22.5 psi\\T_1=110^oC=383.15 K\\P_2=?\\T_2=65.9^oC=339.05 K[/tex]
Putting values in above equation, we get:
[tex]\frac{22.5 psi}{383.15 K}=\frac{p_2}{339.05 K}[/tex]
[tex]P_2=\frac{22.5 psi}{383.15 K}\times 339.05 K=19.91 psi[/tex]
The final pressure at 65.9°C is 19.91 psi.
