States Of Matter

The equation of the graph at constant temperature for an ideal gas follows Boyle's law, which is: Given the graph, it suggests a straight-line relation between pressure and the inverse of volume, which implies that the slope of the line is related to the equation. The correct relation would be: Thus, the slope is .

Step 1: Understand Isobaric Graph:
An isobar is a graph of Volume ( ) versus Temperature ( ) at constant Pressure ( ). From the ideal gas equation: Rearranging for vs : This represents a straight line equation passing through origin, where , , and slope .
Step 2: Calculation:
Given: Slope . Moles . Universal Gas Constant (using atm and L units). Equating slope:
Step 3: Final Answer:
The pressure is 0.1 atm.

Step 1: Write the equation of state for an ideal gas.
The ideal gas law is . We are given that mole, so .

Step 2: Express the relationship for an isobar.
An isobar is a graph plotted at constant pressure ( ). The graph is plotted with Volume ( ) on the y-axis and Temperature ( ) on the x-axis.
We rearrange the ideal gas law to express as a function of :

Step 3: Determine the slope of the isobar.
The equation is a straight line equation of the form , where , , and is the slope. The slope of the isobar is:

Step 4: Relate the slopes to the given pressures.
We are given three different pressures, , and their corresponding slopes .

Step 5: Determine the relationship between the slopes.
We are given the pressure relationship . Since the slope is inversely proportional to the pressure (i.e., ), a smaller pressure corresponds to a larger slope. Therefore, the order of the slopes must be the inverse of the order of the pressures. Multiplying by (which is a positive constant) maintains the inequality:

According to the ideal gas law, . Since temperature is constant, the number of moles is proportional to the product .
Let's find the number of moles (or a quantity proportional to it) for each gas before mixing.
For Oxygen (O ) in flask A: atm. L. Number of moles of O , .
For Helium (He) in flask B: atm. L. Number of moles of He, .
After mixing, the total number of moles is .
The mole fraction of oxygen in the final mixture is given by:
.
We are given that the mole fraction of oxygen is 0.2.
.
Now, we solve this equation for V.
.
.
.
L.
The volume of flask 'A' is 0.3 L.

Step 1: Understanding the Concept:
This problem connects two concepts from the kinetic theory of gases: the root mean square (rms) velocity and the average kinetic energy of gas molecules. The average kinetic energy of a gas depends only on its temperature, while the rms velocity depends on both temperature and molar mass.
Step 2: Key Formula or Approach:
1. Root Mean Square (rms) Velocity: , where is the ideal gas constant, is the absolute temperature, and is the molar mass in kg/mol.
2. Average Kinetic Energy per mole: .
We can see a relationship between these two formulas: from the rms velocity formula, , which means . And from the KE formula, . We can also relate them directly: .
Step 3: Detailed Explanation:
We are given:
- Gas: Argon (Ar)
- Molar mass, (must be in SI units).
- RMS velocity, .
- Temperature = T(K).
We need to find the average kinetic energy per mole ( ) at this same temperature T.
Using the direct relationship derived in Step 2:
Substitute the given values (making sure M is in kg/mol):
Alternative Method (Finding T first):
First, find the temperature T using the rms velocity formula. (Let's use R 8.314 J/mol·K, though it's not needed for the final answer).
Now, calculate the average kinetic energy per mole:
The terms cancel out.
Both methods give the same result.
Step 4: Final Answer:
The average kinetic energy of the gas is 8 J mol⁻¹. Therefore, option (A) is correct.

1. An isobar is a plot at constant pressure; for ideal gas, .
2. For n=1 mole, , so slope of V vs T is .
3. Given slope = 0.082 L K , R = 0.082 L atm mol K .
4. Thus, , so atm.
5. Units confirm: slope in L/K, R/P in (L atm / mol K) / atm = L / mol K, but for n=1, yes.
6. Therefore, the correct option is (3) 1.

• For ideal gas, p V .
• Slope ratio .
• K.