The Question is from neet-chemistry chemical-thermodynamics daltons-law, and it was created by the author Mr. S M HUSSAIN

Question

A mixture of $N_2$ and $O_2$ gases has a total pressure of 1.2 atm.  If the partial pressure of $N_2$ is 0.8 atm and the volume of the container is halved while maintaining a constant temperature, what will be the new partial pressure of $O_2$?

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Dalton's Law of Partial Pressures and Boyle's Law
 **Related Topics**: Ideal Gas Law
 **Prerequisites**: Understanding of partial pressures and the relationship between pressure and volume at constant temperature.
## SOLUTION PATHWAY
**Given Information**:
  Total pressure ($P_{total}$) = 1.2 atm
  Partial pressure of $N_2$ ($P_{N_2}$) = 0.8 atm
  Initial Volume = $V_1$
  Final Volume ($V_2$) = $V_1/2$
  Temperature is constant.
**Method & Steps**:
 1. Find the initial partial pressure of $O_2$ ($P_{O_2}$).
  Reasoning: According to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures of the individual gases. 
  Formula/Rule: $P_{total} = P_{N_2} + P_{O_2}$
 2. Apply Boyle's Law to find the new partial pressure of $O_2$.
  Working:  Boyle's Law states that at constant temperature, the pressure of a gas is inversely proportional to its volume ($P_1V_1 = P_2V_2$). Since the temperature is constant, we can apply Boyle's Law to the $O_2$ gas individually.
  $P_{O_2,initial}V_1 = P_{O_2,final}V_2$
  $P_{O_2,final} = P_{O_2,initial} 	imes \frac{V_1}{V_2} = P_{O_2,initial} 	imes \frac{V_1}{V_1/2} = 2P_{O_2,initial}$
  Key Point: Halving the volume doubles the pressure.
 3. Calculate the new partial pressure of $O_2$.
  From step 1: $P_{O_2,initial} = P_{total} - P_{N_2} = 1.2 \, atm - 0.8 \, atm = 0.4 \, atm$
  Substituting into the equation from step 2:
  $P_{O_2,final} = 2 	imes 0.4 \, atm = 0.8 \, atm$
  Result: The new partial pressure of $O_2$ is 0.8 atm.
  Verification:  The total pressure will double when the volume is halved, becoming 2.4 atm. The partial pressure of $N_2$ will also double to 1.6 atm. Therefore, the partial pressure of $O_2$ will be $2.4 \, atm - 1.6 \, atm = 0.8 \, atm$, confirming our result.
 ## OPTION ANALYSIS
A :x: Incorrect. This is the initial partial pressure of $O_2$.
B ✓: Correct. This is the new partial pressure of $O_2$ after the volume is halved.
C :x: Incorrect. This is the initial partial pressure of $N_2$.
D :x: Incorrect. This is double the total pressure, not the partial pressure of $O_2$.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Forgetting to apply Boyle's Law to find the new partial pressure after the volume change.: Remember that pressure and volume are inversely related at constant temperature.
  2.Adding the partial pressures of N2 and O2 after the volume change incorrectly.: The total pressure will double, but the ratio of partial pressures remains the same.
 **Quick Tips**:
  1.Always start by applying Dalton's Law to find initial partial pressures.
  2.Remember that at constant temperature, halving the volume doubles the pressure and vice versa.

Mr. S M HUSSAIN · Answer

0.2 atm

Mr. S M HUSSAIN · Answer

0.4 atm

Mr. S M HUSSAIN · Answer

The correct Option is "0.8 atm"

Mr. S M HUSSAIN · Answer

1.6 atm

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