# If the quantity of oxygen occupying 2.76L container at a…

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**If the quantity of oxygen occupying 2.76L container at a pressure of 0.825 atm and 300k is reduced by one-half, what is the pressure exerted by the remaining gas?**

**A.**** 1.650atm**

**B.**** 0.825atm**

**C.**** 0.413atm**

**D.**** 0.275atm**

**QUICK ANSWER…**

**C**

**DETAILS…**

**Please observe that the volume of the container and the temperature remained constant, only the quantity (moles) and the pressure are changing. We have a formula in mind:**

**PV = n RT**

**P = n RT/V**

**Therefore, P/n = RT/V**

**P/n = constant**

**Where P = pressure,**

**T = temperature,**

**V = volume,**

**R = gas constant, and**

**n = number of moles also taken as the quantity.**

**Now, since P/n = constant, it implies that P1/n1 = P2/n2**

**Where P1 is given as 0.825atm**

**P2 = unknown = ?**

** Also, let n1 = A**

**And since the quantity was said to have reduced by half, n2 = A/2 =**

**Substituting these values into the equation, we have:**

**P1/n1 = P2/n2**

**0.825/A = P2/(A/2)**

**(0.825/ A) **

**×**

**(**~~A~~/2) = P2

**Therefore, P2 = 0.825/2 = 0.4125 = approximately 0.413atm**

**Now for the right answer to the above question:**

**Option A is incorrect.****Option B is incorrect.****C is correct.****D is not the correct answer.**

**KEY-POINTS…**

**You may please note these/this:**

**There is a short cut, since P and n are directly proportional, a decrease in P will also cause a decrease in n in the same measure.****A decrease in n by half will trigger a decrease in P by the same half, hence, we can quickly see that P2 = half of P1.**

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**/ culled from 2020 JAMB-UTME chemistry past question 13 /**

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