1. A sample of gas initially occupies 2.52 L at a pressure of 0.912 atm at 22.6°C. What will the pressure be if the temperature is changed to 65.5°C, and the volume is changed to 1.70 L?
Ideal and Real Gases
Ideal gases obey conditions of the general gas laws under all states of pressure and temperature. Ideal gases are also named perfect gases. The attributes of ideal gases are as follows,
Gas Laws
Gas laws describe the ways in which volume, temperature, pressure, and other conditions correlate when matter is in a gaseous state. The very first observations about the physical properties of gases was made by Robert Boyle in 1662. Later discoveries were made by Charles, Gay-Lussac, Avogadro, and others. Eventually, these observations were combined to produce the ideal gas law.
Gaseous State
It is well known that matter exists in different forms in our surroundings. There are five known states of matter, such as solids, gases, liquids, plasma and Bose-Einstein condensate. The last two are known newly in the recent days. Thus, the detailed forms of matter studied are solids, gases and liquids. The best example of a substance that is present in different states is water. It is solid ice, gaseous vapor or steam and liquid water depending on the temperature and pressure conditions. This is due to the difference in the intermolecular forces and distances. The occurrence of three different phases is due to the difference in the two major forces, the force which tends to tightly hold molecules i.e., forces of attraction and the disruptive forces obtained from the thermal energy of molecules.
![### Gas Law Application Problem
**Problem Statement:**
A sample of gas initially occupies 2.52 L at a pressure of 0.912 atm and a temperature of 22.6°C. What will the pressure be if the temperature is changed to 65.5°C, and the volume is changed to 1.70 L?
**Solution Approach:**
1. **Understand the problem:**
- Initial volume (\(V_1\)): 2.52 L
- Initial pressure (\(P_1\)): 0.912 atm
- Initial temperature (\(T_1\)): 22.6°C (which needs to be converted to Kelvin)
- Final volume (\(V_2\)): 1.70 L
- Final temperature (\(T_2\)): 65.5°C (which needs to be converted to Kelvin)
2. **Conversion of temperatures to Kelvin:**
\[ T(\text{K}) = T(\text{°C}) + 273.15 \]
\[
T_1(\text{K}) = 22.6 + 273.15 = 295.75 \text{ K}
\]
\[
T_2(\text{K}) = 65.5 + 273.15 = 338.65 \text{ K}
\]
3. **Using the Combined Gas Law formula:**
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Rearrange to solve for \(P_2\):
\[
P_2 = \frac{P_1 V_1 T_2}{T_1 V_2}
\]
Substitute the values into the equation:
\[
P_2 = \frac{(0.912 \text{ atm}) (2.52 \text{ L}) (338.65 \text{ K})}{(295.75 \text{ K}) (1.70 \text{ L})}
\]
Simplify the expression to find \(P_2\).
**Conclusion:**
The calculation will provide the final pressure \(P_2\) in atm, which is the unknown we need to determine.
**Note:** Detailed steps and final calculated value can be](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F878f7b50-6050-483a-af47-6a6f18eb659f%2F8b7884a5-8097-4f21-b212-5e170b3e607e%2Fgn2df9i_processed.jpeg&w=3840&q=75)
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