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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




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 =

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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:

  1. Option A is incorrect.
  2. Option B is incorrect.
  3. C is correct.
  4. D is not the correct answer.


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 /

2 Answers

  1. 0 Votes Thumb up 0 Votes Thumb down 0 Votes

    Feel free to ask us questions on this topic, we will respond in minutes…

    Admin - May 31, 2021 | Reply

  2. 0 Votes Thumb up 0 Votes Thumb down 0 Votes

    it was clear

    ADEBODUN HELLEN - Apr 19, 2022 | Reply

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