application of pressure law (Pressure law)

Question 1

📝 Problem:
A gas initially has a pressure of 1.0 atm at a temperature of 25°C. If the temperature of the gas is increased to 75°C while keeping the volume constant, what will be the new pressure of the gas?

Answer 1

🧠 Solution:
According to Gay-Lussac’s Law (the pressure law), when the volume of a gas is held constant, the pressure of the gas is directly proportional to its absolute temperature.

Given:
- Initial pressure (( P_1 )) = 1.0 atm
- Initial temperature (( T_1 )) = 25°C
- Final temperature (( T_2 )) = 75°C (temperature increased)

First, we need to convert the temperatures to Kelvin since Gay-Lussac’s Law requires temperatures in Kelvin scale:
[ T(K) = T(°C) + 273.15 ]

For initial temperature (( T_1 )):
[ T_1(K) = 25°C + 273.15 = 298.15 \, \text{K} ]

For final temperature (( T_2 )):
[ T_2(K) = 75°C + 273.15 = 348.15 \, \text{K} ]

Now, we can use Gay-Lussac’s Law equation:
[ \frac{P_1}{T_1} = \frac{P_2}{T_2} ]

Substitute the given values:
[ \frac{1.0 \, \text{atm}}{298.15 \, \text{K}} = \frac{P_2}{348.15 \, \text{K}} ]

Cross multiply:
[ 1.0 \times 348.15 = P_2 \times 298.15 ]

[ P_2 = \frac{1.0 \times 348.15}{298.15} ]

[ P_2 \approx 1.17 \, \text{atm} ]

So, the new pressure of the gas (( P_2 )) is approximately 1.17 atm.

✨ Answer:
The new pressure of the gas is approximately 1.17 atm.