Question 10 of 12 The gas in a 250.0 mL piston experiences a change in pressure from 1.00 atm to 3.25 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant?
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.
![**Boyle's Law Application: Calculating New Volume of Gas in a Piston**
**Problem Statement:**
The gas in a 250.0 mL piston experiences a change in pressure from 1.00 atm to 3.25 atm. What is the new volume (in mL) assuming the moles of gas and temperature are held constant?
**Conceptual Explanation:**
This problem can be solved using Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature and moles of gas are constant. Mathematically, it is represented as:
\[ P_1 \times V_1 = P_2 \times V_2 \]
Where:
- \( P_1 \) = initial pressure (1.00 atm)
- \( V_1 \) = initial volume (250.0 mL)
- \( P_2 \) = final pressure (3.25 atm)
- \( V_2 \) = final volume (unknown)
**Solution Steps:**
1. Substitute the known values into Boyle's Law equation:
\[
1.00 \, \text{atm} \times 250.0 \, \text{mL} = 3.25 \, \text{atm} \times V_2
\]
2. Solve for \( V_2 \):
\[
V_2 = \frac{1.00 \, \text{atm} \times 250.0 \, \text{mL}}{3.25 \, \text{atm}}
\]
3. Calculate \( V_2 \):
\[
V_2 = \frac{250.0}{3.25} \approx 76.92 \, \text{mL}
\]
**Conclusion:**
The new volume of the gas in the piston, when the pressure is increased to 3.25 atm, is approximately 76.92 mL. This decrease in volume with an increase in pressure is consistent with the inverse relationship described by Boyle's Law.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa04b93a0-8cb6-4c4e-81fa-ab66edf67455%2F26c14ac2-137d-4ac9-8ef3-3b607476e8bd%2F1yo845_processed.jpeg&w=3840&q=75)
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