IO: 10.0 L of a gas is found t0 exert 97.0 kPa at 25.0 %C. What would be the required temperature (in Celsius) to change the pressure to standard pressure?’
The Correct Answer and Explanation is:
To solve this problem, we can use Gay-Lussac’s Law, which relates the pressure and temperature of a gas when the volume and amount of gas remain constant. The formula is:P1/T1=P2/T2P_1/T_1 = P_2/T_2P1/T1=P2/T2
Where:
- P1P_1P1 is the initial pressure (97.0 kPa),
- T1T_1T1 is the initial temperature in Kelvin,
- P2P_2P2 is the final pressure (standard pressure = 101.3 kPa),
- T2T_2T2 is the final temperature in Kelvin, which we need to find.
Step 1: Convert the initial temperature to Kelvin
The initial temperature is given as 25.0°C. To convert to Kelvin, add 273.15:T1=25.0°C+273.15=298.15 KT_1 = 25.0°C + 273.15 = 298.15 \, \text{K}T1=25.0°C+273.15=298.15K
Step 2: Set up the equation using Gay-Lussac’s Law
Using the formula P1/T1=P2/T2P_1/T_1 = P_2/T_2P1/T1=P2/T2, we substitute the known values:97.0 kPa298.15 K=101.3 kPaT2\frac{97.0 \, \text{kPa}}{298.15 \, \text{K}} = \frac{101.3 \, \text{kPa}}{T_2}298.15K97.0kPa=T2101.3kPa
Step 3: Solve for T2T_2T2
Rearranging the equation to solve for T2T_2T2:T2=101.3 kPa×298.15 K97.0 kPaT_2 = \frac{101.3 \, \text{kPa} \times 298.15 \, \text{K}}{97.0 \, \text{kPa}}T2=97.0kPa101.3kPa×298.15KT2=30272.39597.0≈312.7 KT_2 = \frac{30272.395}{97.0} \approx 312.7 \, \text{K}T2=97.030272.395≈312.7K
Step 4: Convert the final temperature back to Celsius
To convert the final temperature back to Celsius:T2=312.7 K−273.15=39.5°CT_2 = 312.7 \, \text{K} – 273.15 = 39.5°CT2=312.7K−273.15=39.5°C
Final Answer:
The required temperature to change the pressure to standard pressure is 39.5°C.
Explanation:
This problem illustrates how the pressure of a gas is directly proportional to its temperature when the volume is constant. By using Gay-Lussac’s Law, we can predict that increasing the temperature of the gas will increase its pressure, and conversely, decreasing the temperature will lower the pressure. In this case, to reach standard pressure (101.3 kPa), the temperature needs to increase from 25°C to 39.5°C. This demonstrates the relationship between temperature and pressure in gases, assuming no volume change.
